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Answer :
To solve this problem, let's go through the steps needed to find out how many grams of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex] are produced with a 64.13% yield when 3.5 moles of [tex]\( \text{HC}_2\text{H}_3\text{O}_2 \)[/tex] are reacted.
Step 1: Determine the balanced chemical equation.
The given chemical reaction is:
[tex]\[ \text{HC}_2\text{H}_3\text{O}_2 + \text{Sr(OH)}_2 \rightarrow \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 + \text{H}_2\text{O} \][/tex]
This equation is already balanced.
Step 2: Understand the mole-to-mole relationship.
According to the balanced chemical equation, 1 mole of [tex]\( \text{HC}_2\text{H}_3\text{O}_2 \)[/tex] reacts to produce 1 mole of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex].
Step 3: Calculate the molar mass of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex].
- Strontium (Sr): approximately 87.62 g/mol
- Carbon (C, 4 atoms in total): [tex]\( 4 \times 12.01 = 48.04 \)[/tex] g/mol
- Hydrogen (H, 6 atoms in total): [tex]\( 6 \times 1.008 = 6.048 \)[/tex] g/mol
- Oxygen (O, 4 atoms in total): [tex]\( 4 \times 16.00 = 64.00 \)[/tex] g/mol
Adding these together gives us the molar mass of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex]:
[tex]\[ 87.62 + 48.04 + 6.048 + 64.00 = 205.708 \, \text{g/mol} \][/tex]
Step 4: Calculate the theoretical yield in moles.
Given that we started with 3.5 moles of [tex]\( \text{HC}_2\text{H}_3\text{O}_2 \)[/tex], and the reaction produces the same amount of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex], the theoretical yield is 3.5 moles of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex].
Step 5: Calculate the actual moles produced considering the yield.
The actual yield is 64.13% of the theoretical yield:
[tex]\[ \text{Actual yield in moles} = 3.5 \times 0.6413 = 2.24455 \, \text{moles} \][/tex]
Step 6: Convert moles to grams.
To find out how many grams of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex] are actually produced, use the molar mass:
[tex]\[ \text{Mass} = \text{moles} \times \text{molar mass} = 2.24455 \times 205.708 \][/tex]
This gives us approximately 461.722 grams of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex].
So, the final answer is that approximately 461.72 grams of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex] are produced with a 64.13% yield when 3.5 moles of [tex]\( \text{HC}_2\text{H}_3\text{O}_2 \)[/tex] are reacted.
Step 1: Determine the balanced chemical equation.
The given chemical reaction is:
[tex]\[ \text{HC}_2\text{H}_3\text{O}_2 + \text{Sr(OH)}_2 \rightarrow \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 + \text{H}_2\text{O} \][/tex]
This equation is already balanced.
Step 2: Understand the mole-to-mole relationship.
According to the balanced chemical equation, 1 mole of [tex]\( \text{HC}_2\text{H}_3\text{O}_2 \)[/tex] reacts to produce 1 mole of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex].
Step 3: Calculate the molar mass of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex].
- Strontium (Sr): approximately 87.62 g/mol
- Carbon (C, 4 atoms in total): [tex]\( 4 \times 12.01 = 48.04 \)[/tex] g/mol
- Hydrogen (H, 6 atoms in total): [tex]\( 6 \times 1.008 = 6.048 \)[/tex] g/mol
- Oxygen (O, 4 atoms in total): [tex]\( 4 \times 16.00 = 64.00 \)[/tex] g/mol
Adding these together gives us the molar mass of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex]:
[tex]\[ 87.62 + 48.04 + 6.048 + 64.00 = 205.708 \, \text{g/mol} \][/tex]
Step 4: Calculate the theoretical yield in moles.
Given that we started with 3.5 moles of [tex]\( \text{HC}_2\text{H}_3\text{O}_2 \)[/tex], and the reaction produces the same amount of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex], the theoretical yield is 3.5 moles of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex].
Step 5: Calculate the actual moles produced considering the yield.
The actual yield is 64.13% of the theoretical yield:
[tex]\[ \text{Actual yield in moles} = 3.5 \times 0.6413 = 2.24455 \, \text{moles} \][/tex]
Step 6: Convert moles to grams.
To find out how many grams of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex] are actually produced, use the molar mass:
[tex]\[ \text{Mass} = \text{moles} \times \text{molar mass} = 2.24455 \times 205.708 \][/tex]
This gives us approximately 461.722 grams of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex].
So, the final answer is that approximately 461.72 grams of [tex]\( \text{Sr(C}_2\text{H}_3\text{O}_2\text{)}_2 \)[/tex] are produced with a 64.13% yield when 3.5 moles of [tex]\( \text{HC}_2\text{H}_3\text{O}_2 \)[/tex] are reacted.
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