Question

for the following reaction, 19.4 grams of iron are allowed to react with 9.33 grams of oxygen gas. iron (s) + oxygen (g) → iron(ii) oxide (s) what is the maximum amount of iron(ii) oxide that can be formed? grams what is the formula for the limiting reagent? what amount of the excess reagent remains after the reaction is complete? grams

Explanation:

Step1: Balance the chemical equation

$2Fe(s)+O_2(g)
ightarrow 2FeO(s)$

Step2: Calculate moles of reactants

Molar mass of $Fe$ is $55.85\ g/mol$. Moles of $Fe=\frac{19.4\ g}{55.85\ g/mol}\approx0.347\ mol$. Molar mass of $O_2$ is $32\ g/mol$. Moles of $O_2=\frac{9.33\ g}{32\ g/mol}\approx0.292\ mol$.

Step3: Determine the limiting reagent

From the balanced equation, the mole - ratio of $Fe$ to $O_2$ is $2:1$. For $0.292\ mol$ of $O_2$, the moles of $Fe$ required is $2\times0.292 = 0.584\ mol$. But we have only $0.347\ mol$ of $Fe$. So, $Fe$ is the limiting reagent.

Step4: Calculate moles of $FeO$ formed

Since the mole - ratio of $Fe$ to $FeO$ is $1:1$ in the balanced equation, moles of $FeO$ formed = moles of $Fe$ (limiting reagent) = $0.347\ mol$.

Step5: Calculate mass of $FeO$ formed

Molar mass of $FeO$ is $55.85 + 16=71.85\ g/mol$. Mass of $FeO$ formed = $0.347\ mol\times71.85\ g/mol\approx24.9\ g$.

Step6: Calculate moles of $O_2$ used

Moles of $O_2$ used = $\frac{0.347\ mol}{2}=0.1735\ mol$.

Step7: Calculate mass of excess $O_2$

Moles of excess $O_2=0.292\ mol - 0.1735\ mol = 0.1185\ mol$. Mass of excess $O_2=0.1185\ mol\times32\ g/mol\approx3.8\ g$.

Answer:

What is the maximum amount of iron(II) oxide that can be formed? 24.9 grams
What is the FORMULA for the limiting reagent? $Fe$
What amount of the excess reagent remains after the reaction is complete? 3.8 grams