Question

gaseous methane (ch₄) will react with gaseous oxygen (o₂) to produce gaseous carbon dioxide (co₂) and gaseous water (h₂o). suppose 15.2 g of methane is mixed with 27. g of oxygen. calculate the minimum mass of methane that could be left over by the chemical reaction. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Write the balanced chemical equation

$CH_4 + 2O_2
ightarrow CO_2+2H_2O$

Step2: Calculate the molar masses

The molar mass of $CH_4$ is $12.01 + 4\times1.01=16.05$ g/mol, and the molar mass of $O_2$ is $2\times16.00 = 32.00$ g/mol.

Step3: Determine the moles of reactants

The moles of $CH_4$, $n_{CH_4}=\frac{15.2}{16.05}\approx0.947$ mol. The moles of $O_2$, $n_{O_2}=\frac{27}{32.00}= 0.844$ mol.

Step4: Identify the limiting reactant

From the balanced - equation, the mole - ratio of $CH_4$ to $O_2$ is 1:2. For 0.844 mol of $O_2$, the moles of $CH_4$ required is $\frac{0.844}{2}=0.422$ mol.

Step5: Calculate the mass of unreacted $CH_4$

The mass of unreacted $CH_4$ is $(0.947 - 0.422)\times16.05=(0.525)\times16.05 = 8.43$ g.

Answer:

8.4 g