Aluminium carbide reacts with water according to the following equatio

1.	Aluminium carbide reacts with water according to the following equation :Al4C3 + 12H2O ---> 4Al (OH)3 + 3CH4(i) What mass of aluminium hydroxide is formed from 12 g of aluminium carbide ? (ii) What volume of methane at s.t. p. is obtained from 12 g of aluminium carbide ? [Relative molecular weight of Al4C3 = 144; Al(OH)3 = 78]
Answer» (i) Al₄C₃ + 12H₂O ---> 4Al (OH)₃ + 3CH₄ One mole of Al₄C₃ gives 4 moles of Al(OH)₃ i.e., 144 g of Al₄C₃ gives 4 x 78 g of Al(OH)₃ So, 12 g of AI₄C₃ gives 312 x 12/144 g of Al(OH)₃ = 26 g of Al(OH)₃. (ii) One mole of Al₄C₃ gives 3 moles of methane 12 g of Al₄C₃ gives 48 x 12/144 g of CH₄ = 4 g Now, 16 g of methane has volume 22.4 L (at STP, the volume of one mole of any gas is 22.4 L) 4 g of methane would occupy 5.6 L. So, 5.6 L of methane would be obtained from 12 g of Al₄C₃.

Aluminium carbide reacts with water according to the following equation :Al4C3 + 12H2O ---> 4Al (OH)3 + 3CH4(i) What mass of aluminium hydroxide is formed from 12 g of aluminium carbide ? (ii) What volume of methane at s.t. p. is obtained from 12 g of aluminium carbide ? [Relative molecular weight of Al4C3 = 144; Al(OH)3 = 78]

Answer»

(i) Al₄C₃ + 12H₂O ---> 4Al (OH)₃ + 3CH₄

One mole of Al₄C₃ gives 4 moles of Al(OH)₃

i.e., 144 g of Al₄C₃ gives 4 x 78 g of Al(OH)₃

So, 12 g of AI₄C₃ gives 312 x 12/144 g of Al(OH)₃

= 26 g of Al(OH)₃.

(ii) One mole of Al₄C₃ gives 3 moles of methane

12 g of Al₄C₃ gives 48 x 12/144 g of CH₄ = 4 g

Now, 16 g of methane has volume 22.4 L (at STP, the volume of one mole of any gas is 22.4 L)

4 g of methane would occupy 5.6 L.

So, 5.6 L of methane would be obtained from 12 g of Al₄C₃.