If the reaction of 0.112 grams of H2 produces 0.745 grams of H2O, what is the percent yield?
Fe3O4+4H2=3Fe+4H2O

Question

If the reaction of 0.112 grams of H2 produces 0.745 grams of H2O, what is the percent yield?
Fe3O4+4H2=>3Fe+4H2O

Ellie Andrews · Accepted Answer

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 %#

Naomi Aronson · Answer

undefined To find the percent yield:

1. Calculate the theoretical yield of water (H2O) using stoichiometry.
2. Compare the theoretical yield with the actual yield.
3. Calculate the percent yield using the formula:

$$ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% $$

Given:
- Mass of H2 = 0.112 grams
- Mass of H2O = 0.745 grams

Using stoichiometry:

1 mole of H2 produces 4 moles of H2O.

$$ \text{Molar mass of H2O} = 18.015 \, \text{g/mol} $$
$$ \text{Molar mass of H2} = 2.016 \, \text{g/mol} $$

$$ \text{Theoretical Yield of H2O} = \frac{0.112 \, \text{g H2}}{2.016 \, \text{g/mol H2}} \times \frac{4 \, \text{mol H2O}}{1 \, \text{mol H2}} \times 18.015 \, \text{g/mol H2O} $$

$$ \text{Theoretical Yield of H2O} ≈ 4.494 \, \text{g} $$

$$ \text{Percent Yield} = \left( \frac{0.745 \, \text{g}}{4.494 \, \text{g}} \right) \times 100\% ≈ 16.57\% $$

Therefore, the percent yield is approximately 16.57%.

If the reaction of 0.112 grams of H2 produces 0.745 grams of H2O, what is the percent yield? Fe3O4+4H2=&gt;3Fe+4H2O

If the reaction of 0.112 grams of H2 produces 0.745 grams of H2O, what is the percent yield? Fe3O4+4H2=>3Fe+4H2O