Question

Aluminum carbide $\left({\mathbf{Al}}_{\mathbf{4}}{\mathbf{C}}_{\mathbf{3}}\right)$reacts with water to produce gaseous methane $\left({\mathbf{CH}}_{\mathbf{4}}\right)$. Calculate the mass of methane formed from 63.2g $\left({\mathbf{Al}}_{\mathbf{4}}{\mathbf{C}}_{\mathbf{3}}\right)$

Short answer

Mass of methane produce in the reaction is 21.1g

Step-by-step solution

Equation for the reaction

• When, aluminum carbide reacts with water, it produces gaseous methane along with aluminum oxide.
• Balanced equation is

$A{l}_4{C}_3+12H_{2}O\to 3C{H}_4+4Al{\left(OH\right)}_3$

Mass of methane produced

According to balanced equation 1 mole${\mathrm{Al}}_4{\mathrm{C}}_3$ produced 1 mol of $C{H}_4$.

Then, amount of $C{H}_4$produced from 63.2g of ${\mathrm{Al}}_4{\mathrm{C}}_3$is

$$\begin{gathered} Massof=63.2gA{l}_4{C}_3\times \frac{1molA{l}_4{C}_3}{143.95gA{l}_4{C}_3}\times \frac{3molC{H}_4}{1molA{l}_4{C}_3}\times \frac{16.042gC{H}_4}{1molC{H}_4} \\ =21.1gC{H}_4 \end{gathered}$$

Hence, the mass of methane produced in the reaction is$21.1g$