IDNLearn.com provides a reliable platform for finding accurate and timely answers. Our Q&A platform offers reliable and thorough answers to ensure you have the information you need to succeed in any situation.
Sagot :
To determine how many grams of [tex]\( \text{CO}_2 \)[/tex] are formed from 2.6 moles of [tex]\( \text{C}_2 \text{H}_4 \)[/tex], we need to follow a few steps:
1. Determine the molar mass of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] (Ethene):
- The molar mass of Carbon (C) is 12.01 g/mol.
- The molar mass of Hydrogen (H) is 1.01 g/mol.
- Ethene ([tex]\( \text{C}_2 \text{H}_4 \)[/tex]) consists of 2 Carbon atoms and 4 Hydrogen atoms.
So the molar mass of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] is:
[tex]\[ (2 \times 12.01) + (4 \times 1.01) = 24.02 + 4.04 = 28.06 \text{ g/mol} \][/tex]
2. Determine the molar mass of [tex]\( \text{CO}_2 \)[/tex] (Carbon Dioxide):
- The molar mass of Carbon (C) is 12.01 g/mol.
- The molar mass of Oxygen (O) is 16.00 g/mol.
- [tex]\( \text{CO}_2 \)[/tex] consists of 1 Carbon atom and 2 Oxygen atoms.
So the molar mass of [tex]\( \text{CO}_2 \)[/tex] is:
[tex]\[ (1 \times 12.01) + (2 \times 16.00) = 12.01 + 32.00 = 44.01 \text{ g/mol} \][/tex]
3. Understand the stoichiometry of the reaction:
The balanced chemical equation for the combustion of ethene is:
[tex]\[ \text{C}_2 \text{H}_4 + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2 \text{O} \][/tex]
This equation tells us that 1 mole of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] produces 2 moles of [tex]\( \text{CO}_2 \)[/tex].
4. Calculate the moles of [tex]\( \text{CO}_2 \)[/tex] formed from 2.6 moles of [tex]\( \text{C}_2 \text{H}_4 \)[/tex]:
According to the stoichiometry of the reaction, 1 mole of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] produces 2 moles of [tex]\( \text{CO}_2 \)[/tex]. Therefore, 2.6 moles of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] will produce:
[tex]\[ 2.6 \times 2 = 5.2 \text{ moles of } \text{CO}_2 \][/tex]
5. Calculate the grams of [tex]\( \text{CO}_2 \)[/tex] formed:
Using the molar mass of [tex]\( \text{CO}_2 \)[/tex], we can convert moles to grams:
[tex]\[ 5.2 \text{ moles of } \text{CO}_2 \times 44.01 \text{ g/mol} = 228.852 \text{ grams} \][/tex]
Therefore, the mass of [tex]\( \text{CO}_2 \)[/tex] formed from 2.6 moles of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] is:
[tex]\[ \boxed{228.852 \text{ g}} \][/tex]
1. Determine the molar mass of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] (Ethene):
- The molar mass of Carbon (C) is 12.01 g/mol.
- The molar mass of Hydrogen (H) is 1.01 g/mol.
- Ethene ([tex]\( \text{C}_2 \text{H}_4 \)[/tex]) consists of 2 Carbon atoms and 4 Hydrogen atoms.
So the molar mass of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] is:
[tex]\[ (2 \times 12.01) + (4 \times 1.01) = 24.02 + 4.04 = 28.06 \text{ g/mol} \][/tex]
2. Determine the molar mass of [tex]\( \text{CO}_2 \)[/tex] (Carbon Dioxide):
- The molar mass of Carbon (C) is 12.01 g/mol.
- The molar mass of Oxygen (O) is 16.00 g/mol.
- [tex]\( \text{CO}_2 \)[/tex] consists of 1 Carbon atom and 2 Oxygen atoms.
So the molar mass of [tex]\( \text{CO}_2 \)[/tex] is:
[tex]\[ (1 \times 12.01) + (2 \times 16.00) = 12.01 + 32.00 = 44.01 \text{ g/mol} \][/tex]
3. Understand the stoichiometry of the reaction:
The balanced chemical equation for the combustion of ethene is:
[tex]\[ \text{C}_2 \text{H}_4 + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2 \text{O} \][/tex]
This equation tells us that 1 mole of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] produces 2 moles of [tex]\( \text{CO}_2 \)[/tex].
4. Calculate the moles of [tex]\( \text{CO}_2 \)[/tex] formed from 2.6 moles of [tex]\( \text{C}_2 \text{H}_4 \)[/tex]:
According to the stoichiometry of the reaction, 1 mole of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] produces 2 moles of [tex]\( \text{CO}_2 \)[/tex]. Therefore, 2.6 moles of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] will produce:
[tex]\[ 2.6 \times 2 = 5.2 \text{ moles of } \text{CO}_2 \][/tex]
5. Calculate the grams of [tex]\( \text{CO}_2 \)[/tex] formed:
Using the molar mass of [tex]\( \text{CO}_2 \)[/tex], we can convert moles to grams:
[tex]\[ 5.2 \text{ moles of } \text{CO}_2 \times 44.01 \text{ g/mol} = 228.852 \text{ grams} \][/tex]
Therefore, the mass of [tex]\( \text{CO}_2 \)[/tex] formed from 2.6 moles of [tex]\( \text{C}_2 \text{H}_4 \)[/tex] is:
[tex]\[ \boxed{228.852 \text{ g}} \][/tex]
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.
