Question

When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction CaCO3(s)→CaO(s)+CO2(g) What is the mass of calcium carbonate needed to produce 41.0 L of carbon dioxide at STP?

Answer 1

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Answer 2

Given:

Mass of carbon dioxide = 41.0 L

$V_M$ at STP = 22.4 L/mol

To find:

Mass of calcium carbonate ($CaCO_3)$ = ?

Formula to be used:

$m(CaCO)_3 = \frac{V (CO_2) * M(CaCO_3)}{V_M}$

Calculation:

$m(CaCO)_3 = \frac{V (CO_2) * M(CaCO_3)}{V_M}$

$m(CaCO)_3 = \frac{(41.0 * 100)}{22.4}$

$m(CaCO)_3 = \frac{4100}{22.4}$

$m(CaCO)_3 =183 g$

Conclusion:

At STP, to produce 41.0 L of carbon dioxide, 183 g of calcium carbonate is required.