Certainly! To find out how many grams of [tex]\(\text{H}_2\text{O}\)[/tex] are formed from the complete combustion of 0.18 moles of [tex]\(\text{C}_2\text{H}_4\)[/tex] (ethylene), follow these steps:
1. Balanced Chemical Equation:
The balanced chemical equation for the complete combustion of ethylene ([tex]\(\text{C}_2\text{H}_4\)[/tex]) is:
[tex]\[
\text{C}_2\text{H}_4 + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2\text{O}
\][/tex]
From this equation, we can see that 1 mole of [tex]\(\text{C}_2\text{H}_4\)[/tex] yields 2 moles of [tex]\(\text{H}_2\text{O}\)[/tex].
2. Calculate Moles of [tex]\(\text{H}_2\text{O}\)[/tex] Produced:
Since 1 mole of [tex]\(\text{C}_2\text{H}_4\)[/tex] produces 2 moles of [tex]\(\text{H}_2\text{O}\)[/tex], 0.18 moles of [tex]\(\text{C}_2\text{H}_4\)[/tex] would produce:
[tex]\[
0.18 \, \text{mol} \, \text{C}_2\text{H}_4 \times \frac{2 \, \text{mol} \, \text{H}_2\text{O}}{1 \, \text{mol} \, \text{C}_2\text{H}_4} = 0.36 \, \text{mol} \, \text{H}_2\text{O}
\][/tex]
3. Convert Moles of [tex]\(\text{H}_2\text{O}\)[/tex] to Grams:
The molar mass of [tex]\(\text{H}_2\text{O}\)[/tex] (water) is approximately 18.01528 grams per mole. To find the mass of [tex]\(\text{H}_2\text{O}\)[/tex] produced, we use the following calculation:
[tex]\[
\text{mass of } \text{H}_2\text{O} = \text{moles of } \text{H}_2\text{O} \times \text{molar mass of } \text{H}_2\text{O}
\][/tex]
Substituting the values, we get:
[tex]\[
\text{mass of } \text{H}_2\text{O} = 0.36 \, \text{mol} \, \text{H}_2\text{O} \times 18.01528 \, \text{g/mol} \approx 6.4855008 \, \text{g} \, \text{H}_2\text{O}
\][/tex]
Therefore, the number of grams of [tex]\( \text{H}_2\text{O} \)[/tex] formed from 0.18 mol of [tex]\( \text{C}_2\text{H}_4 \)[/tex] is [tex]\( 6.4855008 \, \text{g} \)[/tex].
If required to round to significant figures, based on significant figures from the given data (0.18), we might round the result to two significant figures as [tex]\( 6.49 \)[/tex] grams of [tex]\( \text{H}_2\text{O} \)[/tex].
