What is the mole ratio of $F e_{3} O_{4}$ $Fe_3O_4$ to $F e$ $Fe$ in the equation $3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$ $3Fe + 4H_2O -> Fe_3O_4 + 4H_2$ ?

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Answer 1 · 2016-03-16T23:28:17

$\frac{F e_{3} O_{4}}{F e}$ $(Fe_3O_4)/(Fe)$ is equal to
$\frac{1 m o l}{3 m o l}$ $(1 mol)/(3 mol)$

To determine the mole ratio of chemical equations, generally is to make sure that the equation is balanced following the Law of the Conservation of Mass;
Since the problem presents a balanced equation, all you have to do is to identify which compounds involved to get the right mole ratio;
The number in front of each compound/element are called coefficients which also serves as the number of moles of the elements/compounds present in the chemical equation;
The balanced equation has the following number of moles
$3 m o l F e$ $3 mol Fe$
$4 m o l H_{2} O$ $4 mol H_2O$
$1 m o l F e_{2} O_{4}$ $1 mol Fe_2O_4$
$4 m o l H_{2}$ $4 mol H_2$
Therefore; the mole ratio is $\frac{1 m o l F e_{2} O_{4}}{3 m o l F e}$ $(1 mol Fe_2O_4)/(3 mol Fe)$

What is the mole ratio of Fe3O4Fe_3O_4 to FeFe in the equation 3Fe+4H2O→Fe3O4+4H23Fe + 4H_2O -> Fe_3O_4 + 4H_2?