Question

The reaction between aluminium carbide and water takes place according to the following equation:
$A{l}_4{C}_3+12H_{2}O\to 3C{H}_4+4Al(OH{)}_3$
Calculate the volume fo methane measured at STP released from 14.4 g of aluminum carbide by excess fo water.

Answer 1

According to balance equation
$A{l}_4{C}_3+12H_{2}O\to 3C{H}_4+4Al(OH{)}_3$
Molar mass of aluminium carbide is 144 g
Number of moles of Aluminium carbide $=\frac{14.4g}{144g}=0.1mol$
​1 mol of aluminium carbide produce 3 moles of methane at STP.
1 mole = 22.4 L at STP
Therefore 3 mole = 3 $\times$ 22.4 L = 67.2 L of methane at STP​
Or 1 mole of Aluminium carbide produce 67.2 L of methane at STP
therefore 0.1 mol of Aluminium carbide will produce =
= 6.72 L of methane