Question

Use the reaction below to determine how many liters of H2O(g) result when 2.7 grams of CH4 are combusted in excess oxygen. The temperature after combustion is 380K and the pressure is 1.5 atm.CH4(g) + 2O2(g) > CO2(g) + 2H2O(g)

Answer 1

1) Write the chemical equation.

`CH_4+2O_2\rightarrow CO_2+2H_2O`

2) List the known and unknown quantities.

Sample: 2.7 g CH4.

Temperature: 380 K.

Pressure: 1.5 atm.

3) Convert grams of CH4 to moles of CH4.

The molar mass of CH4 is 16.0425 g/mol.

`mol\text{ }CH_4=2.7\text{ }g\text{ }CH_4*\frac{1\text{ }mol\text{ }CH_4}{16.0425\text{ }g\text{ }CH_4}=0.17\text{ }mol\text{ }CH_4`

4) Moles of H2O produced from 0.17 mol CH4.

The molar ratio between CH4 and H2O is 1 mol CH4: 2 mol H2O.

`mol\text{ }H_2O=0.17\text{ }CH_4*\frac{2\text{ }mol\text{ }H_2O}{1\text{ }mol\text{ }CH_4}=0.34\text{ }H_2O`

5) Volume of H2O produced in the reaction.

5.1- List the known and unknown quantities.

Moles: 0.34 mol H2O.

Temperature: 380 K.

Pressure: 1.5 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

5.2- Set the equation.

`PV=nRT`

5.3- Plug in the known values and solve for V (liters).

`(1.5\text{ }atm)*(V)=(0.34\text{ }mol\text{ }H_2O)*(0.082057\text{ }L*atm*K^(-1)*mol^(-1))*(380\text{ }K)`
`V=\frac{(0.34\text{ }mol\text{ }H_2O)(0.082057\text{ }L*atm*K^(-1)*mol^(-1))(380\text{ }K)}{1.5\text{ }atm}`
`V=7.1\text{ }L`

7.1 L H2O was produced from 2.7 g CH4 in the reaction.