To solve this problem, we need to determine how many grams of [tex]\( H_2 \)[/tex] will react completely with 6.50 moles of [tex]\( CO \)[/tex] based on the given chemical reaction:
[tex]\[ 8 CO + 17 H_2 \rightarrow C_8H_{18} + 8 H_2O \][/tex]
Here's the step-by-step solution:
### Step 1: Determine the stoichiometric relationship
From the balanced chemical equation, we see that 8 moles of [tex]\( CO \)[/tex] react with 17 moles of [tex]\( H_2 \)[/tex].
### Step 2: Calculate the moles of [tex]\( H_2 \)[/tex] needed
Since the stoichiometric ratio between [tex]\( CO \)[/tex] and [tex]\( H_2 \)[/tex] is [tex]\( \frac{17}{8} \)[/tex], we can calculate the moles of [tex]\( H_2 \)[/tex] required for 6.50 moles of [tex]\( CO \)[/tex]:
[tex]\[ \text{moles of } H_2 = 6.50 \text{ moles} \times \frac{17 \text{ moles } H_2}{8 \text{ moles } CO} \][/tex]
### Step 3: Calculate the moles of [tex]\( H_2 \)[/tex]
[tex]\[ \text{moles of } H_2 = 6.50 \times \frac{17}{8} = 6.50 \times 2.125 = 13.8125 \text{ moles } H_2 \][/tex]
### Step 4: Convert moles of [tex]\( H_2 \)[/tex] to grams
The molar mass of [tex]\( H_2 \)[/tex] is approximately 2.02 g/mol.
[tex]\[ \text{grams of } H_2 = \text{moles of } H_2 \times \text{molar mass of } H_2 \][/tex]
### Step 5: Calculate the grams of [tex]\( H_2 \)[/tex]
[tex]\[ \text{grams of } H_2 = 13.8125 \text{ moles} \times 2.02 \text{ g/mol} \][/tex]
[tex]\[ \text{grams of } H_2 = 27.90125 \text{ grams } H_2 \][/tex]
### Step 6: Round the answer to three significant figures
The final answer rounded to three significant figures is:
[tex]\[ 27.9 \text{ grams of } H_2 \][/tex]
So,
[tex]\[ \boxed{27.9} \text{ grams of } H_2 \text{ will react completely with 6.50 moles of } CO. \][/tex]
