Question

for the question(s) that follow, consider the following equation.
2mg + o₂ → 2mgo
the number of moles of mgo produced when 0.20 mole of o₂ reacts completely is
0.10 mole.
0.60 mole.
0.40 mole.
0.80 mole.
0.20 mole.
question 5 (1 point)
a compound contains 40.0g c, 6.71g h, and 53.29g o. the empirical formula of this compound is __________.
cho₂
c₂h₃o₄
c₂h₄o₂
c₂h₂o₄
ch₂o

Explanation:

Step1: Determine mole ratio from balanced equation

From \(2\text{Mg}+\text{O}_2
ightarrow 2\text{MgO}\), mole ratio of \(\text{O}_2:\text{MgO}=1:2\).

Step2: Calculate moles of \(\text{MgO}\)

If \(n(\text{O}_2) = 0.20\) mol, then \(n(\text{MgO})=2\times n(\text{O}_2)\). So \(n(\text{MgO})=2\times0.20 = 0.40\) mol.

Step1: Calculate moles of each element

Molar mass of \(C = 12.01\) g/mol, \(n(C)=\frac{40.0\space g}{12.01\space g/mol}\approx3.33\) mol.
Molar mass of \(H = 1.008\) g/mol, \(n(H)=\frac{6.71\space g}{1.008\space g/mol}\approx6.66\) mol.
Molar mass of \(O = 16.00\) g/mol, \(n(O)=\frac{53.29\space g}{16.00\space g/mol}\approx3.33\) mol.

Step2: Find mole ratio

Divide each mole value by the smallest (\(3.33\)).
\(C:H:O=\frac{3.33}{3.33}:\frac{6.66}{3.33}:\frac{3.33}{3.33}=1:2:1\)

Answer:

0.40 mole.

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