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Some Basic Concepts of Chemistry
Stoichiometry
6 questions from 5 papers, each with its answer.
- questions
- 6
- papers
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- years asked, 2017–2026
- 2 of 10
- of Some Basic Concepts of Chemistry
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Appears in
Concepts tested
- Limiting reagent
- Mass of precipitate
- Moles of gas from decomposition
- moles of H2 for a given amount of NH3
- Percentage purity
- Stoichiometry
All Questions
2026 Re-NEET 1 question
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The amount of carbon dioxide evolved upon complete combustion of 116 g of n-butane is (Given: atomic mass in amu H = 1, C = 12 and O = 16)n-butane is $C_4H_{10}$; $n=\frac{116}{58}=2$ mol. $C_4H_{10}+\frac{13}{2}O_2\rightarrow4CO_2+5H_2O$, so 2 mol butane gives 8 mol $CO_2$. Mass $=8\times44=352$ g.
2019 1 question
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The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is:Haber's process: $N_2(g)+3H_2(g)\rightarrow2NH_3(g)$ 2 moles of $NH_3$ require 3 moles of $H_2$. So 20 moles of $NH_3$ require $\frac{3\times20}{2}=30$ moles of $H_2$.
2015 AIPMT-II 2 questions
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20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (At. Wt.: Mg = 24)$MgCO_3(s) \rightarrow MgO(s) + CO_2(g)$
Moles of $MgCO_3$ if pure = $\frac{20}{84} = 0.238$ mol, which would give $0.238\times 40 = 9.523$ g MgO.
Actual MgO = 8 g
% purity = $\frac{8}{9.523}\times 100 = 84\%$ -
What is the mass of the precipitate formed when 50 mL of 16.9% solution of $AgNO_3$ is mixed with 50 mL of 5.8% NaCl solution? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)50 mL of 16.9% $AgNO_3$ contains 8.45 g = $\frac{8.45}{170} = 0.049$ mol
50 mL of 5.8% NaCl contains 2.9 g = $\frac{2.9}{58.5} = 0.049$ mol
$AgNO_3 + NaCl \rightarrow AgCl + NaNO_3$, so 0.049 mol AgCl forms.
Mass of AgCl = $0.049\times 143.5 \approx 7$ g
2009 AIPMT 1 question
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10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will beReaction: $2H_2 + O_2 \rightarrow 2H_2O$
Moles of $H_2$ = $\frac{10}{2} = 5$ mol; moles of $O_2$ = $\frac{64}{32} = 2$ mol
2 mol of $O_2$ needs $2\times2 = 4$ mol of $H_2$; 5 mol of $H_2$ is available, so $H_2$ is in excess and $O_2$ is the limiting reagent.
1 mol of $O_2$ gives 2 mol of $H_2O$, so 2 mol of $O_2$ gives $2\times2 = 4$ mol of $H_2O$.
Amount of water produced = 4 mol (1 mol of $H_2$ is left over)
2000 AIPMT 1 question
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Volume of $CO_2$ obtained by the complete decomposition of 9.85 g $BaCO_3$ is :$BaCO_3 \xrightarrow{\Delta} BaO + CO_2$
Molar mass of $BaCO_3 = 137 + 12 + 48 = 197$ g/mol
Moles of $BaCO_3 = \dfrac{9.85}{197} = 0.05$ mol, so 0.05 mol of $CO_2$ is formed.
Volume at STP $= 0.05 \times 22.4 = 1.12$ L