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Answer :
Let's solve the problem step-by-step:
1. Identify the Reactants and Products:
- We have the compound [tex]\( C_3H_{18} \)[/tex] reacting with [tex]\( O_2 \)[/tex].
- The products of this reaction are [tex]\( CO_2 \)[/tex] and [tex]\( H_2O \)[/tex].
2. Understand the relationship between [tex]\( C_3H_{18} \)[/tex] and [tex]\( H_2O \)[/tex]:
- To solve this problem, we assume a hypothetical balanced chemical reaction. In this reaction, each mole of [tex]\( C_3H_{18} \)[/tex] generates 9 moles of [tex]\( H_2O \)[/tex].
3. Calculate the Molar Masses:
- Molar Mass of [tex]\( C_3H_{18} \)[/tex]:
[tex]\[
(3 \, \text{moles of C} \times 12.01 \, \text{g/mol}) + (18 \, \text{moles of H} \times 1.008 \, \text{g/mol}) = 54.174 \, \text{g/mol}
\][/tex]
- Molar Mass of [tex]\( H_2O \)[/tex]:
[tex]\[
(2 \, \text{moles of H} \times 1.008 \, \text{g/mol}) + (1 \, \text{mole of O} \times 16.00 \, \text{g/mol}) = 18.016 \, \text{g/mol}
\][/tex]
4. Determine the moles of [tex]\( H_2O \)[/tex] collected:
- Given that 62.3 g of [tex]\( H_2O \)[/tex] is produced, we calculate the number of moles of [tex]\( H_2O \)[/tex]:
[tex]\[
\frac{62.3 \, \text{g}}{18.016 \, \text{g/mol}} \approx 3.46 \, \text{moles of } H_2O
\][/tex]
5. Calculate the moles of [tex]\( C_3H_{18} \)[/tex] reacted:
- From the reaction stoichiometry, 1 mole of [tex]\( C_3H_{18} \)[/tex] produces 9 moles of [tex]\( H_2O \)[/tex]. Therefore, the moles of [tex]\( C_3H_{18} \)[/tex] that reacted is:
[tex]\[
\frac{3.46 \, \text{moles of } H_2O}{9} \approx 0.384 \, \text{moles of } C_3H_{18}
\][/tex]
6. Calculate the mass of [tex]\( C_3H_{18} \)[/tex] reacted:
- Multiply the moles of [tex]\( C_3H_{18} \)[/tex] by its molar mass:
[tex]\[
0.384 \, \text{moles} \times 54.174 \, \text{g/mol} \approx 20.8 \, \text{g}
\][/tex]
According to our calculations, the mass of [tex]\( C_3H_{18} \)[/tex] that was reacted with oxygen is approximately 20.8 grams. The correct answer is:
b) 20.8 g (light blue)
1. Identify the Reactants and Products:
- We have the compound [tex]\( C_3H_{18} \)[/tex] reacting with [tex]\( O_2 \)[/tex].
- The products of this reaction are [tex]\( CO_2 \)[/tex] and [tex]\( H_2O \)[/tex].
2. Understand the relationship between [tex]\( C_3H_{18} \)[/tex] and [tex]\( H_2O \)[/tex]:
- To solve this problem, we assume a hypothetical balanced chemical reaction. In this reaction, each mole of [tex]\( C_3H_{18} \)[/tex] generates 9 moles of [tex]\( H_2O \)[/tex].
3. Calculate the Molar Masses:
- Molar Mass of [tex]\( C_3H_{18} \)[/tex]:
[tex]\[
(3 \, \text{moles of C} \times 12.01 \, \text{g/mol}) + (18 \, \text{moles of H} \times 1.008 \, \text{g/mol}) = 54.174 \, \text{g/mol}
\][/tex]
- Molar Mass of [tex]\( H_2O \)[/tex]:
[tex]\[
(2 \, \text{moles of H} \times 1.008 \, \text{g/mol}) + (1 \, \text{mole of O} \times 16.00 \, \text{g/mol}) = 18.016 \, \text{g/mol}
\][/tex]
4. Determine the moles of [tex]\( H_2O \)[/tex] collected:
- Given that 62.3 g of [tex]\( H_2O \)[/tex] is produced, we calculate the number of moles of [tex]\( H_2O \)[/tex]:
[tex]\[
\frac{62.3 \, \text{g}}{18.016 \, \text{g/mol}} \approx 3.46 \, \text{moles of } H_2O
\][/tex]
5. Calculate the moles of [tex]\( C_3H_{18} \)[/tex] reacted:
- From the reaction stoichiometry, 1 mole of [tex]\( C_3H_{18} \)[/tex] produces 9 moles of [tex]\( H_2O \)[/tex]. Therefore, the moles of [tex]\( C_3H_{18} \)[/tex] that reacted is:
[tex]\[
\frac{3.46 \, \text{moles of } H_2O}{9} \approx 0.384 \, \text{moles of } C_3H_{18}
\][/tex]
6. Calculate the mass of [tex]\( C_3H_{18} \)[/tex] reacted:
- Multiply the moles of [tex]\( C_3H_{18} \)[/tex] by its molar mass:
[tex]\[
0.384 \, \text{moles} \times 54.174 \, \text{g/mol} \approx 20.8 \, \text{g}
\][/tex]
According to our calculations, the mass of [tex]\( C_3H_{18} \)[/tex] that was reacted with oxygen is approximately 20.8 grams. The correct answer is:
b) 20.8 g (light blue)
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