What is the mass of water vapour produced?

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1 Answer
Jan 4, 2018

This problem combines the ideal gas law and stoichiometry t show that 3.78 g of H_2O are produced. The solution follows...

Explanation:

First, we must determine the number of moles of CO_2 produced, because a chemical equation demands all quantities be in moles.

The ideal gas law will help us here:

PV=nRT

where R is the gas constant, 8.314 kPa L/ mol K. We solve for n:

n = (PV)/(RT) = ((100 kPa)(5L))/((8.314)(573 K))=0.105 mol of CO_2

(Note the conversion of Celsius temperature to Kelvin)

Now, we can apply the chemical equation, which tells us that the quantity of H_2O produced is twice the amount of CO2, which we get from the coefficients to the left of these substances.

"mol " H_2O= "mol "CO_2 xx 2 = 0.210 mol H_2O

Since the molar mass of H_2O is 18 g, the mass of H_2O produced is

0.210 "mol" xx 18.0 g/(mol) = 3.78 g