The reaction between aluminium carbide and water takes place according to thefollowing equation:

Al4C3 + 12 H2O 3CH4 + 4Al(OH)3

Calculate the volume of methane measured at STP, released from 14.4 g of aluminium carbide by excess water.

According to balance equation 
Al₄C₃ + 12 H₂O → 3CH₄ + 4Al(OH)3
Molar mass of aluminium carbide is 144 g mol^-1
Number of moles of Aluminium carbide = 14.4 g/ 144 g mol^-1
​                                                                       = 0.1 mol
​1 mol of aluminium carbide produce 3 moles of methane at STP.
1 mole = 22.4 L at STP
Therefore 3 mole = 3 x 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 = 67.2 L/1 mol x 0.1 mol
                                                                                   = 6.72 L of methane