Question

1. the combustion of the organic fuel, decane, is outlined in the chemical equation below. you must balance the equation (coefficients are bigger than you may be used to) in order to answer the subsequent questions a - c.

____c₁₀h₂₂ + o₂ → ____co₂ + ____h₂o

a. how many moles of co₂ are produced if 5.0 moles of c₁₀h₂₂ react with an excess of o₂?

b. how many moles of o₂ react with 0.75 moles of c₁₀h₂₂?

c. how many moles of o₂ would be required to produce 4.0 moles of h₂o?

Explanation:

Step1: Balance the chemical equation

For \(C\) atoms: There are \(10\) \(C\) atoms in \(C_{10}H_{22}\). So the coefficient of \(CO_{2}\) is \(10\).
For \(H\) atoms: There are \(22\) \(H\) atoms in \(C_{10}H_{22}\). So the coefficient of \(H_{2}O\) is \(11\).
For \(O\) atoms: On the right - hand side, there are \(2\times10 + 1\times11=31\) \(O\) atoms. So the coefficient of \(O_{2}\) is \(\frac{31}{2}\). To get rid of the fraction, multiply all coefficients by \(2\). The balanced equation is \(2C_{10}H_{22}+31O_{2}\to20CO_{2}+22H_{2}O\)

Step2: Solve part (a)

From the balanced equation \(2C_{10}H_{22}+31O_{2}\to20CO_{2}+22H_{2}O\), the mole ratio of \(C_{10}H_{22}\) to \(CO_{2}\) is \(2:20 = 1:10\).
If \(n(C_{10}H_{22})=5.0\space mol\), then \(n(CO_{2})=\frac{20}{2}\times5.0\space mol = 50\space mol\)

Step3: Solve part (b)

The mole ratio of \(C_{10}H_{22}\) to \(O_{2}\) is \(2:31\).
If \(n(C_{10}H_{22}) = 0.75\space mol\), then \(n(O_{2})=\frac{31}{2}\times0.75\space mol=11.625\space mol\)

Step4: Solve part (c)

From the balanced equation, the mole ratio of \(H_{2}O\) to \(O_{2}\) is \(22:31\).
If \(n(H_{2}O) = 4.0\space mol\), then \(n(O_{2})=\frac{31}{22}\times4.0\space mol=\frac{124}{22}\space mol\approx5.64\space mol\)

Answer:

a. \(50\space mol\)
b. \(11.625\space mol\)
c. \(\approx5.64\space mol\)