Question

Combining 0.3140.314 mol of Al₄C₃ with excess water produced 96.196.1 g of Al(OH)₃.

Al4C3(s)+12H2O(l)⟶3CH4(g)+4Al(OH)3(s)Al4C3(s)+12H2O(l)⟶3CH4(g)+4Al(OH)3(s)

What is the actual yield of aluminum hydroxide in moles?

What is the theoretical yield of aluminum hydroxide in moles?

What is the percent yield?

Answer 1

1)

Molar mass of Al(OH)3,

MM = 1*MM(Al) + 3*MM(O) + 3*MM(H)

= 1*26.98 + 3*16.0 + 3*1.008

= 78.004 g/mol

mass(Al(OH)3)= 96.1 g

use:

number of mol of Al(OH)3,

n = mass of Al(OH)3/molar mass of Al(OH)3

=(96.1 g)/(78 g/mol)

= 1.232 mol

Answer: 1.23 mol

2)

From reaction,

Mol of Al(OH)3 = 4*mol of Al4C3

= 4*0.314 mol

= 1.256 mol

Answer: 1.26 mol

3)

% yield = actual mol * 100 / theoretical mol

= 1.23 * 100 / 1.26

= 97.6 %

Answer: 97.6 %