Question

How many grams of Fe3O4 are required to react completely with 300 grams of H2? Fe3O4 + H2=Fe + H2O

Answer 1

Answer:

8613.5 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For $H_2$

Mass of $H_2$  = 300 g

Molar mass of $H_2$  = 2.016 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{300\ g}{2.016\ g/mol}$

$Moles\ of\ H_2= 148.81\ mol$

According to the given reaction:

$Fe_3O_4+4H_2\rightarrow 3Fe+4H_2O$

4 moles of $H_2$ will react 1 mole of $Fe_3O_4$

1 mole  of $H_2$ will react 1/4 mole of $Fe_3O_4$

148.81 moles of $H_2$ will react 148.81/4 mole of $Fe_3O_4$

Moles of $Fe_3O_4$ required = 37.2025 mol

Molar mass of $Fe_3O_4$ = 231.53 g/mol

Mass of $Fe_3O_4$ = Moles × Molar mass = 37.2025 × 231.53 g = 8613.5 g

Answer 2

Answer:

8596 g of $Fe_3O_4$ is present

Explanation:

Given the equation is

$\mathrm{Fe}_{3} \mathrm{O}_{4}+\mathrm{H}_{2}=\mathrm{Fe}+\mathrm{H}_{2} \mathrm{O}$

On balancing the equation, we get,

$\mathrm{Fe}_{3} \mathrm{O}_{4}+4 \mathrm{H}_{2}=3 \mathrm{Fe}+4 \mathrm{H}_{2} \mathrm{O}$

So, 1 mole of $Fe_3O_4$ reacts with 4 moles of $H_2$

Therefore the ratio is 1:4

Moles of $H_2$used in the reaction = 300/0.02 = 148.5 moles

$Fe_3O_4$present in the reaction =

$148.5 \times \frac{1}{4} \times 231.55=8596 g$

Therefore, 8596 g of $Fe_3O_4$ is present