Question

Assuming an efficiency of 22.80%, calculate the actual yield of magnesium nitrate formed from 127.6 g of magnesium and excess copper(II) nitrate.

Mg+Cu(NO3)2⟶Mg(NO3)2+Cu

Answer 1

Solution:

Percent yield = Actual yield/Theoretical yield × 100%

from 1 mol of Mg can be produced 1 mol of Mg(NO₃)₂.

molar mass M of Mg 24.3 gram/mol

molar mass M of Mg (NO₃)₂ 148.3 gram/mol

actual yield of Mg(NO₃)₂ $= (m Mg )/(M Mg)×(M Mg(NO)_3 )_2)/1$ × 22.80%/100%

actual yield of Mg(NO₃)₂ = 127.6 g Mg/24.3 g/mol Mg × 148.3 g/mol Mg(NO₃)₂ /1 × 22.80%/100% = 177.55 g

Answer: actual yield of Mg(NO₃)₂ 177.55 grams.