Question

for the question(s) that follow, consider the following equation.
2mg + o₂ → 2mgo
the number of moles of oxygen gas needed to react with 4.0 moles of mg is
1.0 mole.
2.0 moles.
3.0 moles.
4.0 moles.
6.0 moles.
question 2 (1 point)
a compound that is composed of only carbon and hydrogen contains 85.7g c and 14.3 g h. what is the empirical formula of the compound?
ch₄
ch₂
c₄h₈
c₂h₄
c₈₆h₁₄

Explanation:

Step1: Determine mole ratio from reaction stoichiometry

From the balanced equation \(2Mg + O_2
ightarrow 2MgO\), the mole ratio of \(Mg\) to \(O_2\) is \(2:1\).

Step2: Calculate moles of \(O_2\)

Let \(x\) be the moles of \(O_2\). Using the ratio \(\frac{2\space mol\space Mg}{1\space mol\space O_2}=\frac{4.0\space mol\space Mg}{x}\). Cross - multiply: \(2x = 4.0\times1\). Solve for \(x\): \(x=\frac{4.0}{2}=2.0\space mol\).

for Question 2:

Step1: Calculate moles of \(C\) and \(H\)

Molar mass of \(C = 12.01\space g/mol\), moles of \(C=\frac{85.7\space g}{12.01\space g/mol}\approx7.14\space mol\). Molar mass of \(H = 1.008\space g/mol\), moles of \(H=\frac{14.3\space g}{1.008\space g/mol}\approx14.2\space mol\).

Step2: Find the simplest ratio

Divide each mole value by the smaller one (\(n = 7.14\)). For \(C\): \(\frac{7.14}{7.14}=1\). For \(H\): \(\frac{14.2}{7.14}\approx2\).

Answer:

2.0 moles.