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Chapter 3: Problem 168

Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right) :\) $$ \mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If 20.4 g of pentane are burned in excess oxygen, what mass of water can be produced, assuming 100\(\%\) yield?

Short Answer

Expert verified
The mass of water produced when 20.4 g of pentane is burned in excess oxygen with a 100% yield is 26.15 g.

Step by step solution

01

Balance the chemical equation

To balance the chemical equation, we need to make sure that the same number of atoms of each element is present on both sides of the equation. The balanced chemical equation for the combustion of pentane is: $$ \mathrm{C}_{5} \mathrm{H}_{12}(l) + 8\mathrm{O}_{2}(g) \longrightarrow 5\mathrm{CO}_{2}(g) + 6\mathrm{H}_{2} \mathrm{O}(l) $$
02

Convert the mass of pentane to moles

Using the periodic table, we can calculate the molar mass of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) as follows: $$ M_{C_{5}H_{12}} = (5 \times 12.01 \,\text{g/mol}) + (12 \times 1.01 \,\text{g/mol}) = 72.05 + 12.12 = 84.17\, \text{g/mol} $$ Now, we can convert the mass of pentane (20.4 g) to moles: $$ \text{moles of pentane} = \frac{\text{mass of pentane}}{M_{C_{5}H_{12}}} = \frac{20.4\, \text{g}}{84.17\, \text{g/mol}} = 0.242\, \text{mol} $$
03

Apply stoichiometry to calculate the moles of water produced

From the balanced equation, we know that for the combustion of one mole of pentane, six moles of water are produced: $$ \mathrm{C}_{5} \mathrm{H}_{12}(l) \longrightarrow 6\mathrm{H}_{2} \mathrm{O}(l) $$ So, we can calculate the moles of water produced when 0.242 moles of pentane are burned: $$ \text{moles of water} = 0.242\, \text{mol} \times 6 = 1.452\, \text{mol} $$
04

Convert the moles of water to mass

Finally, we can calculate the mass of water produced using its molar mass (18.02 g/mol): $$ \text{mass of water} = \text{moles of water} \times M_{H_{2}O} = 1.452\, \text{mol} \times 18.02\, \text{g/mol} = 26.15\, \text{g} $$ Thus, 26.15 grams of water can be produced when 20.4 grams of pentane are burned in excess oxygen, assuming 100% yield.

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Most popular questions from this chapter

Vitamin A has a molar mass of 286.4 \(\mathrm{g} / \mathrm{mol}\) and a general molecular formula of \(\mathrm{C}_{x} \mathrm{H}, \mathrm{E}\) , where \(\mathrm{E}\) is an unknown element. If vitamin \(\mathrm{A}\) is 83.86$\% \mathrm{C}\( and 10.56\)\% \mathrm{H}$ by mass, what is the molecular formula of vitamin A?

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