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Sagot :
Let's solve each part of this question step-by-step using stoichiometry from the balanced equation of the combustion of ethylene \(\left( \text{C}_2\text{H}_4\right)\):
[tex]\[ \text{C}_2\text{H}_4 + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2\text{O} \][/tex]
Part 1: Grams of \(\text{CO}_2\) formed from 1.9 mol of \(\text{C}_2\text{H}_4\)
1. Identify the molar ratio between \(\text{C}_2\text{H}_4\) and \(\text{CO}_2\):
According to the balanced equation, 1 mole of \(\text{C}_2\text{H}_4\) produces 2 moles of \(\text{CO}_2\).
2. Calculate the moles of \(\text{CO}_2\) formed:
Since 1.9 moles of \(\text{C}_2\text{H}_4\) are given:
[tex]\[ \text{Moles of \(\text{CO}_2\)} = 1.9 \text{ mol \(\text{C}_2\text{H}_4\)} \times 2 \frac{\text{mol \(\text{CO}_2\)}}{\text{mol \(\text{C}_2\text{H}_4\)}} \][/tex]
[tex]\[ \text{Moles of \(\text{CO}_2\)} = 3.8 \text{ mol} \][/tex]
3. Calculate the grams of \(\text{CO}_2\) formed:
The molar mass of \(\text{CO}_2\) (carbon dioxide) is 44.01 g/mol.
[tex]\[ \text{Mass of \(\text{CO}_2\)} = 3.8 \text{ mol} \times 44.01 \frac{\text{g}}{\text{mol}} \][/tex]
[tex]\[ \text{Mass of \(\text{CO}_2\)} = 167.24 \text{ g} \][/tex]
Thus, the number of grams of \(\text{CO}_2\) formed from 1.9 mol of \(\text{C}_2\text{H}_4\) is:
[tex]\[ 167.24 \text{ g CO}_2 \][/tex]
Part 2: Grams of \(\text{H}_2\text{O}\) formed from 0.35 mol of \(\text{C}_2\text{H}_4\)
1. Identify the molar ratio between \(\text{C}_2\text{H}_4\) and \(\text{H}_2\text{O}\):
According to the balanced equation, 1 mole of \(\text{C}_2\text{H}_4\) produces 2 moles of \(\text{H}_2\text{O}\).
2. Calculate the moles of \(\text{H}_2\text{O}\) formed:
Since 0.35 moles of \(\text{C}_2\text{H}_4\) are given:
[tex]\[ \text{Moles of \(\text{H}_2\text{O}\)} = 0.35 \text{ mol} \times 2 \frac{\text{mol \(\text{H}_2\text{O}\)}}{\text{mol \(\text{C}_2\text{H}_4\)}} \][/tex]
[tex]\[ \text{Moles of \(\text{H}_2\text{O}\)} = 0.7 \text{ mol} \][/tex]
3. Calculate the grams of \(\text{H}_2\text{O}\) formed:
The molar mass of \(\text{H}_2\text{O}\) (water) is 18.02 g/mol.
[tex]\[ \text{Mass of \(\text{H}_2\text{O}\)} = 0.7 \text{ mol} \times 18.02 \frac{\text{g}}{\text{mol}} \][/tex]
[tex]\[ \text{Mass of \(\text{H}_2\text{O}\)} = 12.614 \text{ g} \][/tex]
Thus, the number of grams of \(\text{H}_2\text{O}\) formed from 0.35 mol of \(\text{C}_2\text{H}_4\) is:
[tex]\[ 12.614 \text{ g H}_2\text{O} \][/tex]
Thus in grams:
[tex]\[ 167.24 \text{ g CO}_2, \quad 12.614 \text{ g H}_2\text{O} \][/tex]
[tex]\[ \text{C}_2\text{H}_4 + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2\text{O} \][/tex]
Part 1: Grams of \(\text{CO}_2\) formed from 1.9 mol of \(\text{C}_2\text{H}_4\)
1. Identify the molar ratio between \(\text{C}_2\text{H}_4\) and \(\text{CO}_2\):
According to the balanced equation, 1 mole of \(\text{C}_2\text{H}_4\) produces 2 moles of \(\text{CO}_2\).
2. Calculate the moles of \(\text{CO}_2\) formed:
Since 1.9 moles of \(\text{C}_2\text{H}_4\) are given:
[tex]\[ \text{Moles of \(\text{CO}_2\)} = 1.9 \text{ mol \(\text{C}_2\text{H}_4\)} \times 2 \frac{\text{mol \(\text{CO}_2\)}}{\text{mol \(\text{C}_2\text{H}_4\)}} \][/tex]
[tex]\[ \text{Moles of \(\text{CO}_2\)} = 3.8 \text{ mol} \][/tex]
3. Calculate the grams of \(\text{CO}_2\) formed:
The molar mass of \(\text{CO}_2\) (carbon dioxide) is 44.01 g/mol.
[tex]\[ \text{Mass of \(\text{CO}_2\)} = 3.8 \text{ mol} \times 44.01 \frac{\text{g}}{\text{mol}} \][/tex]
[tex]\[ \text{Mass of \(\text{CO}_2\)} = 167.24 \text{ g} \][/tex]
Thus, the number of grams of \(\text{CO}_2\) formed from 1.9 mol of \(\text{C}_2\text{H}_4\) is:
[tex]\[ 167.24 \text{ g CO}_2 \][/tex]
Part 2: Grams of \(\text{H}_2\text{O}\) formed from 0.35 mol of \(\text{C}_2\text{H}_4\)
1. Identify the molar ratio between \(\text{C}_2\text{H}_4\) and \(\text{H}_2\text{O}\):
According to the balanced equation, 1 mole of \(\text{C}_2\text{H}_4\) produces 2 moles of \(\text{H}_2\text{O}\).
2. Calculate the moles of \(\text{H}_2\text{O}\) formed:
Since 0.35 moles of \(\text{C}_2\text{H}_4\) are given:
[tex]\[ \text{Moles of \(\text{H}_2\text{O}\)} = 0.35 \text{ mol} \times 2 \frac{\text{mol \(\text{H}_2\text{O}\)}}{\text{mol \(\text{C}_2\text{H}_4\)}} \][/tex]
[tex]\[ \text{Moles of \(\text{H}_2\text{O}\)} = 0.7 \text{ mol} \][/tex]
3. Calculate the grams of \(\text{H}_2\text{O}\) formed:
The molar mass of \(\text{H}_2\text{O}\) (water) is 18.02 g/mol.
[tex]\[ \text{Mass of \(\text{H}_2\text{O}\)} = 0.7 \text{ mol} \times 18.02 \frac{\text{g}}{\text{mol}} \][/tex]
[tex]\[ \text{Mass of \(\text{H}_2\text{O}\)} = 12.614 \text{ g} \][/tex]
Thus, the number of grams of \(\text{H}_2\text{O}\) formed from 0.35 mol of \(\text{C}_2\text{H}_4\) is:
[tex]\[ 12.614 \text{ g H}_2\text{O} \][/tex]
Thus in grams:
[tex]\[ 167.24 \text{ g CO}_2, \quad 12.614 \text{ g H}_2\text{O} \][/tex]
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