Question

What is the mass of water vapour produced?

Answer 1

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`