Question

for the following equation.
2mg(s) + o₂(g) → 2mgo(s)
how many grams of mgo are produced when 40.0 grams of o2 react completely with mg?
a 50.4 g
b 30.4 g
c 60.8 g
d 101 g

Explanation:

Step1: Calculate the moles of \(O_2\)

The molar mass of \(O_2\) is \(M = 32\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 40.0\space g\) (mass of \(O_2\)) and \(M = 32\space g/mol\) (molar mass of \(O_2\)).
\(n_{O_2}=\frac{40.0\space g}{32\space g/mol}= 1.25\space mol\)

Step2: Determine the mole ratio of \(O_2\) to \(MgO\)

From the balanced chemical equation \(2Mg(s)+O_2(g)\to2MgO(s)\), the mole ratio of \(O_2\) to \(MgO\) is \(1:2\).
So, if \(n_{O_2}=1.25\space mol\), then \(n_{MgO}=2\times n_{O_2}\)
\(n_{MgO}=2\times1.25\space mol = 2.5\space mol\)

Step3: Calculate the mass of \(MgO\)

The molar mass of \(MgO\) is \(M=(24 + 16)\space g/mol=40\space g/mol\)
Using the formula \(m = n\times M\), where \(n = 2.5\space mol\) (moles of \(MgO\)) and \(M = 40\space g/mol\) (molar mass of \(MgO\))
\(m_{MgO}=2.5\space mol\times40\space g/mol = 100\space g\approx101\space g\) (due to rounding differences in molar - mass calculations in some textbooks, if we take more precise molar masses: \(M_{O_2}=31.998\space g/mol\), \(n_{O_2}=\frac{40.0}{31.998}\approx1.25\space mol\), \(M_{MgO}=24.305 + 15.999 = 40.304\space g/mol\), \(m_{MgO}=2\times\frac{40.0}{31.998}\times40.304\approx101\space g\))

Answer:

D. 101 g