Question

Assuming an efficiency of 38.60%, calculate the actual yield of magnesium nitrate formed from 145.9 g of magnesium and excess copper(II) nitrate.
I got 230.5g but it isn't right..
I got 6 mols of Mg then calculated molar mass of Mg(NO3)2 and got 148.31 and then multiplied the two together
Then I used the percent yield given to find the actual yield.

Answer 1

Mg + Cu(NO3)2 ==> Mg(NO3)2 + Cu
mols Mg = 145.9/24.3 = 6.003
The equation is 1 mol to 1 mol; therefore, you will have 6.003 mol Mg(NO3)2 formed. Molar mass I have is 148.31 to give g Mg(NO3)2 =890.29g at 100% yield.
890.29 x 0.3860 = 343.65g actual yield. You're allowed 4 s.f. and I would round to 343.6g.

Answer 2

Thank you!