given the following chemical reaction, calculate the mass of iron (iii) oxide produced, when 6.90 grams of…

Answer:

21.6 g

Explanation:

Step1: Calculate moles of oxygen

$n_{O_2}=\frac{m}{M}=\frac{6.90\ g}{32\ g/mol}= 0.215625\ mol$

Step2: Determine mole - ratio

From the balanced equation $4Fe + 3O_2\rightarrow2Fe_2O_3$, the mole - ratio of $O_2$ to $Fe_2O_3$ is $\frac{n_{Fe_2O_3}}{n_{O_2}}=\frac{2}{3}$.

Step3: Calculate moles of iron(III) oxide

$n_{Fe_2O_3}=\frac{2}{3}n_{O_2}=\frac{2}{3}\times0.215625\ mol = 0.14375\ mol$

Step4: Calculate mass of iron(III) oxide

$m_{Fe_2O_3}=n\times M=0.14375\ mol\times159.69\ g/mol\approx21.6\ g$