Part A) When heated, calcium carbonate decomposes to yield calcium oxide and car

Question

Part A) 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 93.0 L of carbon dioxide at STP? Express your answer with the appropriate units.

Part B) Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is 2C4H10(g)+13O2(g)8CO2(g)+10H2O(l) At 1.00 atm and 23 C, what is the volume of carbon dioxide formed by the combustion of 1.60 g of butane? Express your answer with the appropriate units.

Explanation / Answer

1 mole of CaCO3 gives 1 mole of CO2. 1 mole of any gas occupies 22.4 Liters .

The moles of CO2 produced is 93.0/22.4 =4.15 moles

so CaCO3 moles required is 4.15 moles

1 mole of CaCO3 weighs 100 grams

So, mass of 4.15 moles is 415 grams.

(b)

moles of butane=1.60 g/58 =.1.234 moles

1 mole produces 4 moles of CO2

So, 1.234 moles produce 4.94

1 mole of any gas occupy 22.4 L

4.94 moles occupy =22.4*4.94 ===111 L

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