Question

1. a student completes an analysis of a chemical reaction between magnesium and oxygen according the reaction below 2mg(s) + o₂(g) → 2mgo(s) the mass of oxygen that would be required to completely react 4.00 g of magnesium would be 2.63 10.5 5.27 1.32

Explanation:

Step1: Calculate moles of Mg

The molar mass of Mg is approximately 24.31 g/mol. The number of moles of Mg, $n_{Mg}=\frac{m_{Mg}}{M_{Mg}}$, where $m_{Mg} = 4.00$ g and $M_{Mg}=24.31$ g/mol. So $n_{Mg}=\frac{4.00}{24.31}\text{ mol}\approx0.1645$ mol.

Step2: Determine mole - ratio from the balanced equation

From the balanced equation $2Mg(s)+O_2(g)
ightarrow2MgO(s)$, the mole - ratio of Mg to $O_2$ is 2:1. So the number of moles of $O_2$, $n_{O_2}=\frac{1}{2}n_{Mg}$. Substituting $n_{Mg} = 0.1645$ mol, we get $n_{O_2}=\frac{0.1645}{2}=0.08225$ mol.

Step3: Calculate mass of $O_2$

The molar mass of $O_2$ is 32.00 g/mol. The mass of $O_2$, $m_{O_2}=n_{O_2}\times M_{O_2}$. Substituting $n_{O_2}=0.08225$ mol and $M_{O_2} = 32.00$ g/mol, we get $m_{O_2}=0.08225\times32.00 = 2.632$ g $\approx2.63$ g.

Answer:

2.63