Q.When heated, calcium carbonate decomposes to yield calcium oxide and carbon di

Question

Q.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 83.0L of carbon dioxide at STP?

Q.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 4.00g of butane?

Explanation / Answer

Here,

at STP 1 mole gas occupies 22.4 L volume. If CO2 produced is 83 L, moles = 83/22.4 = 3.71 moles of CO2.

the reaction shows 1mole of CO2 is produced by 1 mole of calcium carbonate. This means moles of calcium carbonate is also 3.71 moles. The molecular weight of Calcium carbonate is 100.09. i.e 1 mole has 100.09g mass. So 3.71 moles must have 3.71 * 100.09g = 370.9 g

So, 370.9g CaCO3 is used.

b. Molecualr mass of butane = 58.12g/mol.We have 4.0g. This means moles = 4/58.12 = 0.0688 mol.

Also, two mole butane gives 8 mole CO2. i.e 1 mole butane gives 4 moles CO2. So, 0.0688 mol butane gives 0.0688*4 = 0.275 mol which ocupies 0.275 * 22.4L = 6.17 L

Hope this helps.

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