Question

gaseous butane (ch₃(ch₂)₂ch₃) reacts with gaseous oxygen gas (o₂) to produce gaseous carbon dioxide (co₂) and gaseous water (h₂o). what is the theoretical yield of water formed from the reaction of 43.0 g of butane and 222. g of oxygen gas? be sure your answer has the correct number of significant digits in it.

Explanation:

Step1: Write the balanced chemical equation

$2C_4H_{10}+13O_2
ightarrow8CO_2 + 10H_2O$

Step2: Calculate the molar mass of butane and oxygen

The molar mass of $C_4H_{10}$ ($M_{C_4H_{10}}$) is $4\times12.01+10\times1.01 = 58.14$ g/mol. The molar mass of $O_2$ ($M_{O_2}$) is $2\times16.00 = 32.00$ g/mol.

Step3: Calculate the number of moles of butane and oxygen

The number of moles of butane ($n_{C_4H_{10}}$) is $\frac{43.0}{58.14}\approx0.740$ mol. The number of moles of oxygen ($n_{O_2}$) is $\frac{222}{32.00}= 6.9375$ mol.

Step4: Determine the limiting reactant

From the balanced - equation, the mole - ratio of $C_4H_{10}$ to $O_2$ is $\frac{2}{13}$. For 0.740 mol of $C_4H_{10}$, the moles of $O_2$ required is $0.740\times\frac{13}{2}=4.81$ mol. Since $4.81$ mol < 6.9375 mol, butane is the limiting reactant.

Step5: Calculate the moles of water produced

The mole - ratio of $C_4H_{10}$ to $H_2O$ is $\frac{2}{10}=\frac{1}{5}$. So the moles of water produced ($n_{H_2O}$) is $0.740\times5 = 3.70$ mol.

Step6: Calculate the mass of water produced

The molar mass of water ($M_{H_2O}$) is $2\times1.01+16.00 = 18.02$ g/mol. The mass of water produced ($m_{H_2O}$) is $3.70\times18.02 = 66.7$ g.

Answer:

66.7 g