Chemistry - Chemical Equations and Stoichiometry
Exam Duration: 45 Mins Total Questions : 30
What is the equivalent weight of KMnO4 in alkaline medium?
- (a)
31.60
- (b)
71.00
- (c)
52.66
- (d)
158.00
Change in oxidation state of Mn is from +7 to +6 in alkaline medium.
\(Equivalent\quad mass=\frac { molecular\quad mass }{ change\quad in\quad oxidation\quad state } =\frac { 158 }{ 1 } =158\)
In the following reaction,
\({ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightarrow 2{ NH }_{ 3 }(g)\)
Calculate the moles of NH3 obtained when 2 moles of N2 react with 3 moles of H2 .
- (a)
2 mol
- (b)
4 mol
- (c)
1 mol
- (d)
6 mol
H2 is the limiting reagent. Therefore, 2 moles of NH3 will be obtained.
How many kilograms of CaCO3 will be required to produce 56 kg CaO, if the sample is 90% pure?
- (a)
90 kg
- (b)
110 kg
- (c)
100 kg
- (d)
111.11 kg
\(\underset { 100g }{ { CaCO }_{ 3 } } \rightarrow \underset { 56g }{ Cao } +C{ O }_{ 2 }CaC{ O }_{ 3 }\rightarrow 90%\quad pure\\ \quad \quad \)
56 kg of CaO will be obtained from CaCO3
\(=\frac { 100 }{ 90 } \times 100=111.11\quad kg\)
Calculate the mass per cent of different elements in sodium sulphate.
- (a)
Mass % of Na = 45.0; S = 32.37; O = 22.57
- (b)
Mass % of Na = 32.37; S = 22.57; O = 45.0
- (c)
Mass % of Na = 22.57; S = 45.0; O = 32.37
- (d)
Mass % of Na = 32.37; S =45.0; O = 22.57
Mass of Na2SO4 = 142.04g
Mass % of element= \(atomic\ mass\ of\ element\over molecular\ mass\ of\ compound\times100\)
Mass % of Na=\({46\times100\over 142.04}=32.37\)
Mass % of S = \({32\times100\over 142.04}=22.57\)
Mass % of O=\({64\times100\over 142.04}=45.0\)
What is the percentage of gold in 20 carat gold?
- (a)
80
- (b)
95
- (c)
38.56
- (d)
83.33
24 carats gold = 100% pure
20 carats gold = \(\frac { 100 }{ 24 } \times 20=83.33%\)
2 moles of BaCl2 react with 2 mole of Na3PO4 according to the reaction \(3BaC{ l }_{ 2 }+2N{ a }_{ 3 }{ PO }_{ 4 }\rightarrow { Ba }_{ 3 }{ \left( { PO }_{ 4 } \right) }_{ 2 }+6NaCl\) Calculate the maximum amount of Ba3(PO4)2 obtained.
- (a)
0.13 mol
- (b)
0.55 mol
- (c)
0.24 mol
- (d)
0.667 mol
BaCl2 is the limiting reagent.
\(3\ moles\ of\ BaC{ l }_{ 2 }\equiv 1\ mole\ of\ { Ba }_{ 3 }{ \left( { PO }_{ 4 } \right) }_{ 2 }\\ 2\ moles of\ BaC{ l }_{ 2 }\equiv \frac { 2 }{ 3 } mole\ of \ { Ba }_{ 3 }{ \left( { PO }_{ 4 } \right) }_{ 2 }\ \ \ \ \ =0.667 \ mole \ { Ba }_{ 3 }{ \left( { PO }_{ 4 } \right) }_{ 2 }\quad \)
If 0.56 g of a gas occupies 280 cm3 at STP, then what is the molecular mass of that gas?
- (a)
4.8
- (b)
44.8
- (c)
2
- (d)
22.4
280 cm3 gas at STP weigh = 0.56 g
22400 cm3 gas at STP weigh = \(\frac { 0.56 }{ 280 } \times 22400=44.8\quad g\)
How much of oxygen (in L) is required for the complete oxidation of 1.5 moles of sulphur into sulphur dioxide at STP?
- (a)
11.2
- (b)
22.4
- (c)
33.6
- (d)
44.8
\(S+{ O }_{ 2 }\rightarrow { SO }_{ 2 }\)
1 mole of S requires = 22.4 L of O2
1.5 moles of S requires = 22.4X1.5=33.6 L of O2
If 500 mL of a 5 M solution is diluted to 1500 mL, what will be the molarity of the solution obtained?
- (a)
1.5 M
- (b)
1.66 M
- (c)
0.017 M
- (d)
1.59 M
\({ M }_{ 1 }{ V }_{ 1 }={ M }_{ 2 }{ V }_{ 2 },5\times 500={ M }_{ 2 }\times 1500\\ \Rightarrow { M }_{ 2 }=\frac { 5\times 500 }{ 1500 } =1.66\quad M\)
What will be the molarity of a solution which contains 5.85 g NaCl(s) per 500 mL?
- (a)
4 mol L-1
- (b)
20 mol L-1
- (c)
0.2 mol L-1
- (d)
2 mol L-1
\(Molarity=\frac { 5.85\times 1000 }{ 58.5\times 500 } =0.2\quad mol\quad { L }^{ -1 }\)
One mole of any substance contains 6.022X1023 molecules. Number of molecules of H2SO4 present in 100 mL of 0.02 MH2SO4 solution is
- (a)
12.044X1020 molecules
- (b)
6.022X1023 molecules
- (c)
1X1023 molecules
- (d)
12.044X1023 molecules
\(Number \ of \ moles=\frac { 0.02\times 100 }{ 1000 } =0.002\\ Number \ of \ molecules=0.002\times 6.023\times { 10 }^{ 23 }\)
=12.044 x 1020 molecules
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the following reaction \({ CaCO }_{ 3 }(s)+2HCl(aq)\rightarrow { CaCl }_{ 2 }(aq)+{ CO }_{ 2 }(g)+{ H }_{ 2 }O(l)\) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
- (a)
0.94 g
- (b)
0.24 g
- (c)
0.56 g
- (d)
0.49 g
\(M(HCl)={w\times1000\over m\times V(mL)}⇒0.75={w\times1000\over 36.5\times25}⇒w=0.6844g\)
\(\underset{100g}{CaCO_3}(s)+\underset{73g}{2HCl(aq)}⟶CaCl_2+CO_2+H_2O\)
∵ 73g HCl react with 100g CaCO3
∴ 0.6844g HCl will react with \({100\times0.6844\over 73}\)
=0.9375 ≈ 0.94g CaCO3
\({ 2MnO }_{ 4 }^{ - }+{ 5C }_{ 2 }{ O }_{ 4 }^{ 2- }+{ 16H }^{ + }\rightarrow { 2Mn }^{ 2+ }+{ 10CO }_{ 2 }+{ 8H }_{ 2 }O\) Here, 20 mL of 0.1 M KMnO4 is equivalent to
- (a)
20 mL of 0.5 M H2C2O4
- (b)
50 mL of 0.1 M H2C2O4
- (c)
50 mL of 0.01 M H2C2O4
- (d)
20 mL of 0.1 M H2C2O4
Milliequivalents of
20mL of 0.1M KMnO4=\({20\times0.1\over 2}=1.0\)
20mL of 0.5M H2C2O4=\({20\times0.5\over 5}=2.0\)
50mL of 0.01M H2C2O4=\({20\times0.1\over 5}=0.1\)
20mL of 0.1M H2C2O4=\({20\times0.1\over 5}=0.4\)
Hence., 20mL of 0.1M KMnO4=50 mL of 0.1 M H2C2O4
Calculate the volume ( in mL) of 0.1 M KMnO4 solution required to completely oxidise 100 mL of 0.5 M ferrous sulphate solution in acidic medium?
- (a)
20
- (b)
50
- (c)
200
- (d)
100
Balanced chemical equation
\(MnO_4^-+8H^++5e^e\rightarrow Mn^{2+}+4H_2O\)
\((FeSO_4\rightarrow Fe^{3+}+e^{-})\times5\)
\(⇒\ MnO_4^-≡5Fe^{2+}\)
\({M_1V_1\over n_1}(KMnO_4)={M_2V-2\over n_2}(FeSO_4),{0.1\times V_1\over 1}={0.5\times100\over 5}\)
∴ Volume of KmnO4 = 100mL
How much of 0.1 N HCl is required to react with 1g of CaCO3 ? (Ca = 40, C = 14, O = 16)
- (a)
100 cm3
- (b)
150 cm3
- (c)
250 cm3
- (d)
200 cm3
\(\underset{100g}{CaCO_3}+\underset{2HCl}{2\times36.5g}\rightarrow CaCl_2+H_2O+CO_2\)
∵ 100g of CaCO3 require =2x36.5g of HCl
∴ 1g of CaCO3 requires=\({2\times36.5\over 100}\times1=0.73g\ of\ HCl\)
∴ Amount of HCl present in 0.1N HCl = 36.5 x 0.1
=3.65g
3.65g of HCl contained in 1000mL of 0.1N HCl solution
∴ 0.73g of HCl contained in \({1000\over 3.65}\times0.73=200cm^3of HCl\)
Match the following and choose the correct option.
Column I | Column II | ||
A. | Molarity (M) | p. | Temperature |
B. | Molarity (m) | q. | Dilution |
C. | Mole fraction \((\chi )\) | r. | Volume |
D. | Normality(N) |
- (a)
A B C D p,q,r q,r r q,p - (b)
A B C D p,q,r q q p,q,r - (c)
A B C D q,r p r p,r - (d)
A B C D p,q q q,r p,q
\(A\rightarrow p,q,r\quad B\rightarrow q\quad C\rightarrow q\quad D\rightarrow p,q,r\)
Mixture containing SiO2 and Fe2O3 in 2 g of sample which on very strong heating leave a residue of 1.96 g. The reaction for the lose of weight is \({ 3Fe }_{ 2 }{ O }_{ 3 }(s)\rightarrow { 2Fe }_{ 3 }{ O }_{ 4 }(s)+\frac { 1 }{ 2 } { O }_{ 2 }(g);\) How much of SiO2 is present in original sample?
- (a)
1.2 g
- (b)
0.8 g
- (c)
0.2 g
- (d)
2.0 g
\({ 3Fe }_{ 2 }{ O }_{ 3 }(s)\rightarrow { 2Fe }_{ 3 }{ O }_{ 4 }(s)+\frac { 1 }{ 2 } { O }_{ 2 }\\ \because 16g{ O }_{ 2 }\equiv 480\quad g\quad { Fe }_{ 2 }{ O }_{ 3 }\\ \because 0.04g{ O }_{ 2 }\equiv 0.04\times \frac { 480 }{ 16 } =1.2\quad g\quad { Fe }_{ 2 }{ O }_{ 3 }\)
SiO2 = 2 - 1.2 = 0.8 g
Calculate the mass per cent of calcium, phosphorus and oxygen in calcium phosphate, Ca3(PO4)2 .
- (a)
Ca:38.71%;P:20%;O:41.29%
- (b)
Ca:20%;P:38.7%;O:41.29%
- (c)
Ca:38.7%;P:41.29%O:20%
- (d)
Ca:20%;P:41.29%;O:38.71%
Ca3(PO4)2=310u
Mass % of Ca=\({3\times atomic\ mass\ of\ Ca\over mol.\ mass\ of\ Ca_3(PO_4)_2}\times100\)
\(={3\times40\over 310}\times100=38.71\%\)
Mass % of P=\({2\times atomic\ mass\ of\ P\over mol.\ mass\ of\ Ca_3(PO_4)_2}\times100\)
\(={2\times31\over 310}\times100=20\%\)
Mass % of O=\({8\times atomic\ mass\ of\ P\over mass\ of\ Ca_3(PO_4)_2}\times100\)
\(={8\times16\over 310}\times100=41.29\%\)
A sample of drinking water was found to be severely contaminated with CHCl3 supported to be carcinogenic in nature. The level of contamination was 15ppm. Which of the following is correct?
- (a)
Molarity of CHCl3 in water sample =1.26X10-4 M
- (b)
per cent by mass =1.5X10-2
- (c)
per cent by mass =1.5X10-6
- (d)
Molarity of CHCl3 in water sample = 1.26X10-3 M
15ppm means 15parts in 1 million (106 parts)
% by mass = \({15\times100\over 10^6}=1.5\times10^{-3}\%\)
Molecular mass of CHCl3 = 119g mol-1
1.5 x 10-3 % means 1.5 x 10-3g CHCl3 is present in 100g sample
\(M={w\times1000\over m\times volume\ of\ sample}\)
\(M={1.5\times10^{-3}\times1000\over 119\times100}\)
=0.000126 = 1.26 x 10-4M
In the reaction, \(2A+4B\rightarrow 3C+4D;\) when 5 moles of A react with 6 moles of B, then calculate amount of C formed.
- (a)
10
- (b)
4.5
- (c)
3
- (d)
4
4 moles of B produce = 3 moles of C
\(6\quad moles\quad of\quad B\quad will\quad produce=\frac { 3\times 6 }{ 4 } =4.5\quad moles\quad of\quad C\)
What is the concentration of sugar (C12H22O11) in mol L-1 if its 20 g are dissolved in enough water to make a final volume upto 2 L?
- (a)
0.0292 M
- (b)
0.0120 M
- (c)
0.0391 M
- (d)
0.0200 M
Molar mass of sugar,
\({ C }_{ 12 }{ H }_{ 22 }{ O }_{ 11 }=12\times 12+22\times 1+11\times 16=342g{ mol }^{ -1 }\\ Molarity=\frac { 20 }{ 342\times 2 } =0.0292\quad mol\quad { L }^{ -1 }\quad =0.0292\quad M\)
If the density of methanol is 0.793 kg L-1. What is volume needed for making 2.5 L of its 0.25 M solution?
- (a)
25.22 L
- (b)
2500 mL
- (c)
25.22 mL
- (d)
2500 L
Density = 0.793kgL-1 = 0.793 x 103L-1
Molar mass of methanol
CH3OH = 12 x 1 + 1 x 3 + 1 x 16 + 1 x 1 =32
Molarity=\({0.793\times10^3\over 32g}=24.781M\)
M1V1 = M2V2
24.781 x V1 = 0.25 x 2.5
\(V_1={0.25\times2.5\over 24.781}=0.02522L=25.22mL\)
Calculate the molarity of solution of ethanol in water in which the mole fraction of ethanol is 0.040.
- (a)
2.110 M
- (b)
2.314 M
- (c)
2.000 M
- (d)
2.210 M
1L of ethanol solution = 1L of H2O
Numer of moles of H2O = \({1000\over 18}=55.55mol\)
\(x_{H_2O}={n_{H_2P}\over n_{H_2O}+n_{C_2H_5OH}}\)
\(0.96={55.55\over 55.55+n_{C_2H_5OH}}\)
\(n_{C_2H_5OH}=2.3145mol\)
Molarity is the number of moles of solute (ethanol) in I L solution. So, molarity of ethanol = 2.314 M
Calculate the mass of sodium acetate required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 mol-1 .
- (a)
14.21 g
- (b)
14.71 g
- (c)
15.00 g
- (d)
15.38 g
\(Molarity=\frac { mass\quad of\quad solute\times 1000 }{ molar\quad mass\quad of\quad solute\times volume\quad of\quad solution(in\quad mL) } \\ 0.375=\frac { w\times 1000 }{ 82.0245\times 500 } \Rightarrow w=15.379g\simeq 15.38g\)
2.76 g of silver carbonate on being strongly heated yields residue weighing
- (a)
3.54 g
- (b)
3.0 g
- (c)
1.36 g
- (d)
2.16 g
\(\underset{276g}{Ag_2CO_3}\xrightarrow{\Delta}\underset{216}{2Ag}+CO_2\uparrow+{1\over 2}O_2\uparrow\)
∵ 276g of Ag2CO3 will give 216g of Ag
∴ 2.76g of Ag2CO3 will give \({2.476\times216\over 276}=2.16g\)
Experimentally, it was found that a metal oxide has formula Mo.98O. Metal M, present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be
- (a)
7.01%
- (b)
4.08%
- (c)
6.05%
- (d)
5.08%
M0.98 O.Consider one mole of the oxide
Moles of M (M2+ and M3+ ) = 0.98
Moles of O2- = 1
Let moles of M2+ = x and moles of M3+ = 0.98 - x
On balancing charge, (0.98 - x) x 2+ 3x = 0
1.96 - 2x + 3x - 2 = 0 ⇒ x = 0.04
Percentage of M3+ = \({0.04\over 0.98}\times100=4.08\%\)
The density (in g mL-1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (98 g mol-1) by mass will be
- (a)
1.64
- (b)
1.88
- (c)
1.22
- (d)
1.45
\(Molarity=\frac { 10\times density\times wt.\quad of\quad solute }{ mol.wt.\quad of\quad the\quad solute }\)
\(Density,d=\frac { 3.60\times 98 }{ 10\times 29 } =1.22g{ L }^{ -1 }\)
What volume of hydrogen gas, at 273 K and 1 atm pressure, will be consumed to obtain 21.6 g of elemental boron from the reduction of boron trichloride by hydrogen?
- (a)
89.6 L
- (b)
67.2 L
- (c)
44.8 L
- (d)
22.4 L
\(\underset{2mol}{2BCl_3}+\underset{3mol}{3H_2}\rightarrow\underset{2mol=216g}{2B}+6HCl\)
21.6gB=2mol B ≡ 3mol H2
pV = nRT
\(∴\ v={nRT\over p}={3\times0.0821\times273\over1}=67.2L\)