To find out how many grams of [tex]\( H_2 \)[/tex] will react completely with 6.50 moles of [tex]\( CO \)[/tex], we need to carefully follow the steps of stoichiometry based on the balanced chemical equation:
[tex]\[ 8 \, CO + 17 \, H_2 \rightarrow C_8H_{18} + 8 \, H_2O \][/tex]
Step 1: Determine the molar ratio between [tex]\( CO \)[/tex] and [tex]\( H_2 \)[/tex]:
The balanced equation provides the molar ratio of [tex]\( CO \)[/tex] to [tex]\( H_2 \)[/tex]. From the equation, 8 moles of [tex]\( CO \)[/tex] react with 17 moles of [tex]\( H_2 \)[/tex]. The molar ratio is therefore:
[tex]\[ \frac{17 \, \text{moles of} \, H_2}{8 \, \text{moles of} \, CO} \][/tex]
Step 2: Calculate the moles of [tex]\( H_2 \)[/tex] required:
Given 6.50 moles of [tex]\( CO \)[/tex], we use the molar ratio to find the moles of [tex]\( H_2 \)[/tex] needed:
[tex]\[ \text{Moles of} \, H_2 = 6.50 \, \text{moles of} \, CO \times \frac{17 \, \text{moles of} \, H_2}{8 \, \text{moles of} \, CO} = 13.8125 \, \text{moles of} \, H_2 \][/tex]
Step 3: Convert moles of [tex]\( H_2 \)[/tex] to grams:
The molar mass of [tex]\( H_2 \)[/tex] is approximately 2.016 grams per mole. To get the grams of [tex]\( H_2 \)[/tex], we multiply the moles of [tex]\( H_2 \)[/tex] by the molar mass:
[tex]\[ \text{Grams of} \, H_2 = 13.8125 \, \text{moles} \times 2.016 \, \text{g/mole} = 27.846 \, \text{grams} \][/tex]
Step 4: Round to three significant figures:
Rounding 27.846 grams to three significant figures gives:
[tex]\[ 27.846 \approx 27.8 \, \text{grams} \][/tex]
Hence, the correct answer is:
[tex]\[ \boxed{27.8} \text{ grams of } H_2 \text{ will react completely with } 6.50 \text{ moles of } CO\text{.} \][/tex]
