CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l) Mass of 1 mol = 100 g Mass of 1 mol = 164 g 2.00 g of calcium carbonate (CaCO₃) was reacted with 200 cm³ of 0.1 mol l⁻¹ nitric acid (HNO₃). Take the volume of 1 mole of carbon dioxide to be 24 litres. In the reaction A CaCO₃ is the limiting reactant B an excess of 0.1 mol of nitric acid remains at the end of the reaction C 1 g of calcium nitrate is produced by the reaction D 480 cm³ of carbon dioxide is produced by the reaction.

Scottish Highers Chemistry Getting the most from Reactants: CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(

CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l) Mass of 1 mol = 100 g Mass of 1 mol = 164 g 2.00 g of calcium carbonate (CaCO₃) was reacted with 200 cm³ of 0.1 mol l⁻¹ nitric acid (HNO₃) - Scottish Highers Chemistry - Question 13 - 2016

Question 13

CaCO₃(s)-+-2HNO₃(aq)-→-Ca(NO₃)₂(aq)-+-CO₂(g)-+-H₂O(l)--Mass-of-1-mol-=-100-g-Mass-of-1-mol-=-164-g--2.00-g-of-calcium-carbonate-(CaCO₃)-was-reacted-with-200-cm³-of-0.1-mol-l⁻¹-nitric-acid-(HNO₃)-Scottish Highers Chemistry-Question 13-2016.png

CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l) Mass of 1 mol = 100 g Mass of 1 mol = 164 g 2.00 g of calcium carbonate (CaCO₃) was reacted with 200 cm³ of 0... show full transcript

Worked Solution & Example Answer:CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l) Mass of 1 mol = 100 g Mass of 1 mol = 164 g 2.00 g of calcium carbonate (CaCO₃) was reacted with 200 cm³ of 0.1 mol l⁻¹ nitric acid (HNO₃) - Scottish Highers Chemistry - Question 13 - 2016

Step 1

A CaCO₃ is the limiting reactant

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Answer

To determine if CaCO₃ is the limiting reactant, we first need to calculate the moles of CaCO₃ and HNO₃ involved in the reaction.

Moles of CaCO₃ = $\text{mass} \div \text{molar mass} = \frac{2.00 \text{ g}}{100 \text{ g/mol}} = 0.02 \text{ mol}$

Moles of HNO₃ = $0.1 \text{ mol/l} \times 0.2 \text{ l} = 0.02 \text{ mol}$

From the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of HNO₃. Thus, 0.02 moles of CaCO₃ requires: $0.02 \text{ mol} \times 2 = 0.04 \text{ mol of HNO}_3$

Since we only have 0.02 mol of HNO₃, CaCO₃ is indeed the limiting reactant.

Step 2

B an excess of 0.1 mol of nitric acid remains at the end of the reaction

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Answer

Since we determined that CaCO₃ is limiting, and it requires 0.04 moles of HNO₃ to fully react. However, we only have 0.02 moles of HNO₃ provided. Thus, there is no excess of nitric acid remaining; rather, it is limiting.

Step 3

C 1 g of calcium nitrate is produced by the reaction

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Answer

Based on the balanced equation, 1 mole of CaCO₃ produces 1 mole of Ca(NO₃)₂. Therefore, from 0.02 moles of CaCO₃:

Moles of Ca(NO₃)₂ produced = 0.02 mol Molar mass of Ca(NO₃)₂ = 164 g/mol

Mass of Ca(NO₃)₂ produced = $0.02 \text{ mol} \times 164 \text{ g/mol} = 3.28 \text{ g}$

This means 1 g is not produced; hence this statement is false.

Step 4

D 480 cm³ of carbon dioxide is produced by the reaction

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Answer

As per the stoichiometry of the balanced equation, 1 mole of CaCO₃ produces 1 mole of CO₂.

Moles of CO₂ produced from 0.02 moles of CaCO₃ = 0.02 moles. The volume of CO₂ produced can be calculated as follows:

$0.02 \text{ mol} \times 24 \text{ l/mol} = 0.48 \text{ l} = 480 ext{ cm}^3$

Thus, 480 cm³ of carbon dioxide is produced.

CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l) Mass of 1 mol = 100 g Mass of 1 mol = 164 g 2.00 g of calcium carbonate (CaCO₃) was reacted with 200 cm³ of 0.1 mol l⁻¹ nitric acid (HNO₃) - Scottish Highers Chemistry - Question 13 - 2016

Worked Solution & Example Answer:CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l) Mass of 1 mol = 100 g Mass of 1 mol = 164 g 2.00 g of calcium carbonate (CaCO₃) was reacted with 200 cm³ of 0.1 mol l⁻¹ nitric acid (HNO₃) - Scottish Highers Chemistry - Question 13 - 2016

A CaCO₃ is the limiting reactant

B an excess of 0.1 mol of nitric acid remains at the end of the reaction

C 1 g of calcium nitrate is produced by the reaction

D 480 cm³ of carbon dioxide is produced by the reaction

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