Question

part a
for the reaction, calculate how many grams of the product form when 3.0 g of mg completely reacts.
assume that there is more than enough of the other reactant.
express your answer using two significant figures.
2 mg(s) + o₂(g) → 2 mgo(s)
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Explanation:

Step1: Calculate moles of Mg

Molar mass of Mg is \(24.31\space g/mol\). Moles of Mg, \(n_{Mg}=\frac{mass}{molar\space mass}=\frac{3.0\space g}{24.31\space g/mol}\approx0.1234\space mol\).

Step2: Relate moles of Mg to MgO

From the reaction \(2Mg + O_2
ightarrow2MgO\), the mole ratio of \(Mg:MgO = 2:2 = 1:1\). So moles of \(MgO\), \(n_{MgO}=n_{Mg}\approx0.1234\space mol\).

Step3: Calculate mass of MgO

Molar mass of \(MgO = 24.31 + 16.00 = 40.31\space g/mol\). Mass of \(MgO\), \(m_{MgO}=n_{MgO}\times molar\space mass = 0.1234\space mol\times40.31\space g/mol\approx4.97\space g\). Rounding to two significant figures, \(m_{MgO}\approx5.0\space g\).

Answer:

\(5.0\space g\)