Question

Aluminium carbide reacts with water according to the following equation.
$A{l_4}{C_3} + 12{H_2}O \to 4Al{\left( {OH} \right)_3} + 3C{H_4}$
(i) What mass of aluminium hydroxide is formed from 12g of aluminium carbide?
(II) What volume of methane at s.t.p. is obtained from 12g of aluminium carbide?

Answer 1

Hint: Try to recall that 1 mole of any substance at standard temperature and pressure (s.t.p) occupies volume of 22.4L. Now by using this you can easily answer the given question.

Complete step by step solution:
Given, mass of aluminium carbide, $A{l_4}{C_3}$ = 12g
Molar mass of aluminium carbide,$A{l_4}{C_3} = (27 \text{x} 4) + (12 \times 3)$
$\begin{gathered} = 108 + 36 = 144{\text{ g/mol}} \end{gathered}$
So, number of mole of aluminium carbide,$A{l_4}{C_3}$= $\dfrac{{{\text{mass of aluminium carbide}}}}{{molar{\text{ }}mass{\text{ of aluminium carbide}}}}$
$= \dfrac{{12}}{{144}}$
$= 0.083$.
In reaction, $A{l_4}{C_3} + 12{H_2}O \to 4Al{\left( {OH} \right)_3} + 3C{H_4}$:
1 mole of aluminium carbide,$A{l_4}{C_3}$ on reaction with water to form 4 moles of aluminium hydroxide, $Al{\left( {OH} \right)_3}$.
So, 0.083 mole of aluminium carbide,$A{l_4}{C_3}$ on reaction with water forms $= 0.083 \times 4 = 0.33$ mole of aluminium hydroxide, $Al{\left( {OH} \right)_3}$.
Also, molar mass of aluminium hydroxide, $Al{\left( {OH} \right)_3}$ $= (27) + 3 \times (16 + 1)$
$\begin{gathered} = 27 + 51 = 78{\text{ g/mol}} \end{gathered}$
Hence, mass of aluminium hydroxide, $Al{\left( {OH} \right)_3}$ formed = $78 \times 0.33$
= 25.74 g.
From first part you got, number of moles of aluminium carbide,$A{l_4}{C_3}$= 0.083
In reaction, $A{l_4}{C_3} + 12{H_2}O \to 4Al{\left( {OH} \right)_3} + 3C{H_4}$:
1 mole of aluminium carbide,$A{l_4}{C_3}$on reaction with water forms 3 mole of methane, $C{H_4}$.
So, 0.083 mole of aluminium carbide,$A{l_4}{C_3}$ on reaction with water forms $= 0.083 \times 4 = 2.49$ mole of methane, $C{H_4}$.
Also, it is known to you that 1 mole of any substance occupies 22.4L of volume at standard temperature and pressure i.e. s.t.p.
Hence, 2.49 mole of methane, $C{H_4}$ occupies $= 2.49 \times 22.4 = 55.7$ L of volume at s.t.p.

Therefore, from above we conclude that it forms 25.74 g mass of aluminium hydroxide, $Al{\left( {OH} \right)_3}$ and 55.77 L of methane at s.t.p.

Note:It should be remembered to you that aluminium carbide is used as an abrasive in cutting tools.
Also, you should remember that when magnesium carbide ($M{g_2}C$) reacts with water then it forms ethene gas and magnesium hydroxide.