The percent yield is 74.4 %.
Fe₃O₄ + 4H₂ → 3Fe + 4H₂O
First, calculate the theoretical yield of H₂O.
Theor. yield =
0.112 g H
2
×
1 mol H
2
2.016 g H
2
×
4 mol H
2
O
4 mol H
2
×
18.02 g H
2
O
1 mol H
2
O
=
1.001 g H
2
O
Now calculate the percent yield.
% yield =
actual yield
theoretical yield
×
100
%
=
0.745 g
1.001 g
×
100
%
=
74.4
%
The mass in grams (theoretical yield) of hydrogen is produced is considered to be 0.119 g H2.
Since
Mass of Mg = 31.3 g
Mass of H2O = 2.12 g
First, we have to determine the mole
So,
Mole of Mg = mass of Mg / molar mass of Mg
= (31.3 g ) / (24.30 g/mol )
= 1.28 mole
And,
Mole of H2O = (2.12g ) / (18.01g /mol )
= 0.117 mole
As per the chemical reaction:
2 mole of H2O react with 1 mole of Mg
So, 0.117 mole of H2O will react with mole of Mg is
Now
= ( 1 mol Mg / 2 mol H2O ) * 0.117 mol H2O
= 0.058 mole of Mg
Now
1 mole of Mg produce , 1 mole of H2
So,
0.058 mole of Mg will produce 0.058 mole of H2
Now, determine the mass of H2
= Molar mass of H2 * mole of H2
= 2 * 0.058
= 0.116 g
= 0.119
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