Question

Aluminum carbide reacts with water according to the following equation ${\mathrm{Al}}_4{\mathrm{C}}_3+12{\mathrm{H}}_2\mathrm{O}\to 4\mathrm{Al}{\left(\mathrm{OH}\right)}_3+3{\mathrm{CH}}_4$

(I) What mass of Aluminum hydroxide is formed from 12 g of Aluminum carbide

(ii) What volume of methane at STP is obtained from 12 g of Aluminum carbide relative to molecular weight of Al₄C₃=144 Al(OH)₃=78

Answer 1

Step -1 Calculate mass of Aluminium carbide

$$\begin{gathered} {\mathrm{Al}}_4{\mathrm{C}}_3\left(\mathrm{s}\right)+12{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 4\mathrm{Al}{\left(\mathrm{OH}\right)}_3\left(\mathrm{s}\right)+3{\mathrm{CH}}_4\left(\mathrm{g}\right) \\ \mathrm{Aluminium}\mathrm{Water}\mathrm{Aluminium}\mathrm{hydroxide}\mathrm{Methane} \\ \mathrm{carbide} \end{gathered}$$

1 mole of ${\mathrm{Al}}_4{\mathrm{C}}_3$ produce = 4moles of $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$

The molar mass of $1$mole ${\mathrm{Al}}_4{\mathrm{C}}_3$=144g

The molar mass of$4$moles of $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$ =$312g$

$144g$ of ${\mathrm{Al}}_4{\mathrm{C}}_3$ produce =$312g$of $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$

$1g$ ${\mathrm{Al}}_4{\mathrm{C}}_3$of produce =$\frac{312}{144}$ of $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$

$12g$${\mathrm{Al}}_4{\mathrm{C}}_3$ of produce = $\frac{\left(312\times 12\right)}{144}$ of $\mathrm{Al}{\left(\mathrm{OH}\right)}_3$$=26g$

Step-2 Calculate the Volume of Methane with Aluminium carbide

As per reaction,$1$mole of ${\mathrm{Al}}_4{\mathrm{C}}_3$ gives $3$moles of ${\mathrm{CH}}_4$

$144g$ of ${\mathrm{Al}}_4{\mathrm{C}}_3$ will produce= $3\times 22.4\mathrm{L}\mathrm{of}{\mathrm{CH}}_4$

$12g$ of ${\mathrm{Al}}_4{\mathrm{C}}_3$ will produce= $\left(\frac{3\times 12\times 22.4}{144}\right)=5.6\mathrm{L}{\mathrm{CH}}_4$

Hence,

26g of Aluminum hydroxide is formed.

5.6 lt of methane is obtained.