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Determine the mass (in grams) of [tex]C_4H_{10}[/tex] that is required to produce [tex]14.7 \, \text{g} \, CO_2[/tex]. The balanced equation for the complete combustion of butane is shown below:

[tex]\[
2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O
\][/tex]

- Your answer should have three significant figures.

Provide your answer below:
[tex]\[ \boxed{\phantom{answer}} \][/tex]


Sagot :

To determine the mass of [tex]\( C_4H_{10} \)[/tex] required to produce 14.7 grams of [tex]\( CO_2 \)[/tex], we will follow a step-by-step approach:

1. Calculate the moles of [tex]\( CO_2 \)[/tex] produced:
We start by determining the number of moles of [tex]\( CO_2 \)[/tex] given the mass of [tex]\( CO_2 \)[/tex].

- The molar mass of [tex]\( CO_2 \)[/tex] is 44.01 g/mol.
[tex]\[ \text{Moles of } CO_2 = \frac{\text{mass of } CO_2}{\text{molar mass of } CO_2} = \frac{14.7 \text{ g}}{44.01 \text{ g/mol}} \approx 0.334 \text{ moles} \][/tex]

2. Use stoichiometry to relate moles of [tex]\( CO_2 \)[/tex] to moles of [tex]\( C_4H_{10} \)[/tex]:
From the balanced chemical equation:
[tex]\[ 2 C_4H_{10} + 13 O_2 \rightarrow 8 CO_2 + 10 H_2O \][/tex]
We see that 2 moles of [tex]\( C_4H_{10} \)[/tex] produce 8 moles of [tex]\( CO_2 \)[/tex]. Using this ratio:
[tex]\[ \text{Moles of } C_4H_{10} = \left(\frac{2 \text{ moles of } C_4H_{10}}{8 \text{ moles of } CO_2}\right) \times 0.334 \text{ moles of } CO_2 \approx 0.084 \text{ moles} \][/tex]

3. Convert moles of [tex]\( C_4H_{10} \)[/tex] to mass:
Finally, we convert the moles of [tex]\( C_4H_{10} \)[/tex] to mass in grams.

- The molar mass of [tex]\( C_4H_{10} \)[/tex] is 58.12 g/mol.
[tex]\[ \text{Mass of } C_4H_{10} = \text{moles of } C_4H_{10} \times \text{molar mass of } C_4H_{10} = 0.084 \text{ moles} \times 58.12 \text{ g/mol} \approx 4.853 \text{ grams} \][/tex]

Thus, the mass of [tex]\( C_4H_{10} \)[/tex] required to produce 14.7 grams of [tex]\( CO_2 \)[/tex] is [tex]\( \boxed{4.853} \text{ grams} \)[/tex].