13. what is the limiting reactant if 32.2 g of mg is reacted with 2.5 l of oxygen at stp? how much excess…

Explanation:

Step1: Calculate moles of Mg

The molar - mass of Mg is $M_{Mg}=24.31\ g/mol$. Given $m = 32.2\ g$ of Mg, the number of moles of Mg, $n_{Mg}=\frac{m}{M}=\frac{32.2\ g}{24.31\ g/mol}\approx1.32\ mol$.

Step2: Calculate moles of $O_2$ at STP

At STP, 1 mole of any gas occupies 22.4 L. Given $V = 2.5\ L$ of $O_2$, the number of moles of $O_2$, $n_{O_2}=\frac{V}{V_m}=\frac{2.5\ L}{22.4\ L/mol}\approx0.112\ mol$.

Step3: Determine the limiting reactant

From the balanced chemical equation $2Mg + O_2\rightarrow2MgO$, the mole - ratio of $Mg$ to $O_2$ is 2:1. The moles of Mg required to react with $0.112\ mol$ of $O_2$ is $n_{Mg\ required}=2\times n_{O_2}=2\times0.112\ mol = 0.224\ mol$. Since $n_{Mg}=1.32\ mol$ and $n_{Mg\ required}=0.224\ mol$, $O_2$ is the limiting reactant.

Step4: Calculate the moles of excess reactant (Mg) left

The moles of Mg reacted is $n_{Mg\ reacted}=2\times n_{O_2}=0.224\ mol$. The moles of Mg left, $n_{Mg\ left}=n_{Mg}-n_{Mg\ reacted}=1.32\ mol - 0.224\ mol = 1.096\ mol$. The mass of Mg left, $m_{Mg\ left}=n_{Mg\ left}\times M_{Mg}=1.096\ mol\times24.31\ g/mol\approx26.6\ g$.

Answer:

26.6 g