At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
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]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
