Chemistry, asked by rajeshsharma33, 8 months ago

20 gram of calcium carbonate is react with 20 gram of HCL how much carbon dioxide is produced during this reaction

Answers

Answered by HarshChaudhary0706

Answer:

Balanced reaction: \mathrm{CaCO}_{3}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_{2}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}

From the reaction,

1 mole of calcium carbonate reacts with 2 moles of Hydrogen chloride and produce 1 mole of carbon dioxide.

Let’s calculate the moles of \mathrm{CaCO}_{3}¬:

Molar mass of \mathrm{CaCO}_{3} = 100 g/mol

Given mass of \mathrm{CaCO}_{3}= 20 g

Moles of \mathrm{CaCO}_{3}=\frac{\text { Mass of } \mathrm{CaCo}_{3}}{\text { Molar mass of } \mathrm{CaCo}_{3}}=\frac{20}{100}=0.20\ \mathrm{mol}

0.20 mol of \bold{\mathrm{CaCO}_{3}} reacts with 0.4 mol of HCl

Let’s calculate the moles of HCl:

Molar mass of HCl = 36.45 g/mol

Given mass = 20 ml

Moles of HCl= \frac{\text { Mass of }\mathrm{HCl}}{\text {Molar mass of }\mathrm{HCl}}=\frac{20}{36.45}=0.548\ \mathrm{mol}

Therefore, \mathrm{CaCO}_{3} is limiting reagent.

0.20 mol of \mathrm{CaCO}_{3} gives 0.20 mol of \mathrm{CO}_{2}

Molar mass of \mathrm{CO}_{2} = 44 g/mol

Mass of \mathrm{CO}_{2}\ produced = 0.2 \times 44 \frac{g}{m o l}=8.8 g

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anasuyab46

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Answer:

Here the moles of CaCO3 = Mass of CaCO3/

Molecular Mass

n= 20/100

n= 0.2

Moles of HCl = 20/36.5 = 0.54 mol

The reaction is

CaCO3 + 2HCl = CaCl2 + CO2 + H2O

For 2 mol of HCl, 1 mol CaCO3 is req.

This means for 0.54 mol HCl, 0.27 mol CaCO3 would be req.

But only 0.2 mol CaCO3 is present.

This means CaCO3 is Limiting Reagent.

For 1 mol CaCO3, 1 mol CO2 is formed.

Therefore moles of CO2 produced = 0.27

Mass of CO2 = 44 × 0.27

= 11.88 gm

Explanation: