Answer to Question #239414 in General Chemistry for Simon

Question

The carbon in coal can be converted into methane, CH4, by first heating the coal powder with steam and oxygen, followed by heating with carbon monoxide and hydrogen. The overall process is summarized below.

C(s) + 2 H2(g)  →  CH4(g)

When 10.0 kg of coal is used in the process, 4.20 kg of methane is produced. What is the percentage yield of methane, assuming the coal is pure carbon?

Answer 1

The entire process

$C_{(s)} + O_{2(g)}+ H_2O_{(g)} \to CH_{4(g)}+CO_{2{(g)}}$

The summarised process

$C_{(s)}+ 2 H_{2(g)} → CH_{4(g)}$

From the question;

10kg of coal = $\dfrac{10,000\textsf{ g}}{12\textsf{ g/mol}}$ = 833.3 mol

4.20kg of methane = $\dfrac{4,200\textsf{ g}}{16\textsf{ g/mol}}$ = 262 mol

From the chemical reaction,

1 mole of coal should produce 1 mole of methane

so, 262 moles of methane should be produced by 262 moles of coal

and 833.3 moles of coal should produce 833.3 moles of methane

$\therefore \% \textsf{ yield} = \dfrac{262}{833.3}× 100\%= 31. 441\%$