Question

Assuming a yield of 32.1% 32.1% , calculate the actual yield of magnesium nitrate in grams formed from 148.5 g 148.5 g of magnesium and excess copper(II) nitrate. Mg+Cu ( NO 3 ) 2 ⟶Mg ( NO 3 ) 2 +Cu

actual yield of Mg(NO3)2Mg(NO3)2 : ____g

Answer 1

Given that

Magnesium= 148.5 g and copper in excess, thus the limiting agent is Mg

Mg+Cu ( NO 3 ) 2 --------- > Mg ( NO 3 ) 2 +Cu

Number of mole s= amount in g / molar mass

= 148.5 g/ 24.305 g/ mole

= 6.11 moles Mg

Now calculate the mole of Mg ( NO 3 ) 2 :

6.11 moles Mg * 1 mole Mg ( NO 3 ) 2 /1 mole Mg

= 6.11 mole Mg ( NO 3 ) 2

Amount in g = molar mass * number of molel s

= 148.3 g/mol *6.11

= 906.113 g formed or theoretical yield

% efficiency = actual yield / theoretical yield *100; therefore:

32.1%= actual yield / 906.113 *100

actual yield= 290.86 g