Question

gaseous methane (ch₄) will react with gaseous oxygen (o₂) to produce gaseous carbon dioxide (co₂) and gaseous water (h₂o). suppose 13 g of methane is mixed with 26.6 g of oxygen. calculate the minimum mass of methane that could be left over by the chemical reaction. round your answer to 2 significant digits.

Explanation:

Step1: Write the balanced chemical equation

$$\ce{CH4 + 2O2->CO2 + 2H2O}$$

Step2: Calculate the molar masses

The molar mass of \(CH_4\) is \(M_{CH_4}=(12 + 4\times1)\space g/mol = 16\space g/mol\).
The molar mass of \(O_2\) is \(M_{O_2}=2\times16\space g/mol = 32\space g/mol\).

Step3: Calculate the number of moles of \(O_2\)

The number of moles of \(O_2\), \(n_{O_2}=\frac{m_{O_2}}{M_{O_2}}\)
Given \(m_{O_2} = 26.6\space g\), so \(n_{O_2}=\frac{26.6\space g}{32\space g/mol}= 0.83125\space mol\)

Step4: Determine the moles of \(CH_4\) that react with \(O_2\)

From the balanced equation, the mole ratio of \(CH_4:O_2 = 1:2\).
Let the moles of \(CH_4\) that react be \(n_{CH_4 - react}\). Then \(n_{CH_4 - react}=\frac{n_{O_2}}{2}\)
Substituting \(n_{O_2}=0.83125\space mol\), we get \(n_{CH_4 - react}=\frac{0.83125\space mol}{2}=0.415625\space mol\)

Step5: Calculate the mass of \(CH_4\) that reacts

The mass of \(CH_4\) that reacts, \(m_{CH_4 - react}=n_{CH_4 - react}\times M_{CH_4}\)
Substituting \(n_{CH_4 - react}=0.415625\space mol\) and \(M_{CH_4}=16\space g/mol\), we get \(m_{CH_4 - react}=0.415625\space mol\times16\space g/mol = 6.65\space g\)

Step6: Calculate the mass of \(CH_4\) left over

The initial mass of \(CH_4\) is \(m_{CH_4 - initial}=13\space g\)
The mass of \(CH_4\) left over, \(m_{CH_4 - left}=m_{CH_4 - initial}-m_{CH_4 - react}\)
\(m_{CH_4 - left}=13\space g - 6.65\space g=6.35\space g\approx6.4\space g\)

Answer:

\(6.4\space g\)