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Using the balanced equation for the combustion of octane \(\left( C_8 H_{18} \right)\), answer the following questions.

[tex]\[ 2 C_8 H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2 O \][/tex]

Be sure each of your answer entries has the correct number of significant figures.

Part 1 of 2

How many grams of \( CO_2 \) are formed from \( 2.6 \, \text{mol} \) of \( C_8 H_{18} \) ?

[tex]\[ \square \, \text{g} \, CO_2 \, \square \, \square \times 10 \][/tex]

Part 2 of 2

How many grams of \( H_2 O \) are formed from \( 0.57 \, \text{mol} \) of \( C_8 H_{18} \) ?

[tex]\[ \square \, \text{g} \, H_2 O \, \square \, \square \times 10 \][/tex]



Answer :

GinnyAnswer
Let's solve this problem step-by-step.

## Part 1: Grams of \( CO_2 \) produced from \( 2.6 \) moles of \( C_8H_{18} \)

Using the balanced chemical equation:
[tex]\[ 2 C_8H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2O \][/tex]

We note that 2 moles of \( C_8H_{18} \) produce 16 moles of \( CO_2 \).

Thus,
[tex]\[ \text{Moles of } CO_2 = 2.6 \text{ moles of } C_8H_{18} \times \frac{16 \text{ moles of } CO_2}{2 \text{ moles of } C_8H_{18}} = 2.6 \times 8 = 20.8 \text{ moles of } CO_2 \][/tex]

Next, we calculate the grams of \( CO_2 \) produced. The molar mass of \( CO_2 \) is:
[tex]\[ 12.01 \text{ (C)} + 2 \times 16 \text{ (O)} = 44.01 \text{ g/mol} \][/tex]

Therefore,
[tex]\[ \text{Grams of } CO_2 = 20.8 \text{ moles of } CO_2 \times 44.01 \text{ g/mol} = 915.408 \text{ g} \][/tex]

So,
[tex]\[ \boxed{9.15408} \times 10^2 \text{ g CO}_2 \][/tex]

## Part 2: Grams of \( H_2O \) produced from \( 0.57 \) moles of \( C_8H_{18} \)

Using the balanced chemical equation:
[tex]\[ 2 C_8H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2O \][/tex]

We note that 2 moles of \( C_8H_{18} \) produce 18 moles of \( H_2O \).

Thus,
[tex]\[ \text{Moles of } H_2O = 0.57 \text{ moles of } C_8H_{18} \times \frac{18 \text{ moles of } H_2O}{2 \text{ moles of } C_8H_{18}} = 0.57 \times 9 = 5.13 \text{ moles of } H_2O \][/tex]

Next, we calculate the grams of \( H_2O \) produced. The molar mass of \( H_2O \) is:
[tex]\[ 2 \times 1.008 \text{ (H)} + 16 \text{ (O)} = 18.016 \text{ g/mol} \][/tex]

Therefore,
[tex]\[ \text{Grams of } H_2O = 5.13 \text{ moles of } H_2O \times 18.016 \text{ g/mol} = 92.42208 \text{ g} \][/tex]

So,
[tex]\[ \boxed{9.24221} \times 10 \text{ g H}_2O \][/tex]

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