(Balanced) Equation- 2NaOH + FeCl2 (react to produce) 2NaCl + Fe(OH)2 Question- If 12.75g of NaOH and 20.20g of FeCl2 reacted to produce 18.63g of NaCl, how much Fe(OH)2 was produced?
As they would theoretically produce the same amount, they are all limiting reactants so any can be used for the next step.
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First, we need to determine the limiting reactant to find out how much Fe(OH)2 was produced.
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Calculate the molar masses of the compounds involved:
- NaOH: ( 22.99 , \text{g/mol} + 15.999 , \text{g/mol} + 1.008 , \text{g/mol} = 39.997 , \text{g/mol} )
- FeCl2: ( 55.845 , \text{g/mol} + 2(35.453) , \text{g/mol} = 126.751 , \text{g/mol} )
- NaCl: ( 22.99 , \text{g/mol} + 35.453 , \text{g/mol} = 58.443 , \text{g/mol} )
- Fe(OH)2: ( 55.845 , \text{g/mol} + 2(15.999) , \text{g/mol} + 2(1.008) , \text{g/mol} = 89.847 , \text{g/mol} )
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Convert the given masses to moles:
- Moles of NaOH: ( \frac{12.75 , \text{g}}{39.997 , \text{g/mol}} = 0.3188 , \text{mol} )
- Moles of FeCl2: ( \frac{20.20 , \text{g}}{126.751 , \text{g/mol}} = 0.1595 , \text{mol} )
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Determine the stoichiometric ratio between NaOH and FeCl2 in the balanced equation:
- From the balanced equation, we see that 2 moles of NaOH react with 1 mole of FeCl2.
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Calculate the theoretical yield of NaCl:
- Since 2 moles of NaOH produce 2 moles of NaCl, the theoretical yield of NaCl is ( 2 \times 0.3188 , \text{mol} = 0.6376 , \text{mol} ).
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Determine the actual yield of NaCl:
- Given: 18.63 g of NaCl
- Moles of NaCl: ( \frac{18.63 , \text{g}}{58.443 , \text{g/mol}} = 0.3188 , \text{mol} ) (by comparison to the moles of NaOH)
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Since both reactants produce the same amount of NaCl (0.3188 mol), we conclude that NaOH is the limiting reactant.
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Use the stoichiometry from the balanced equation to find the amount of Fe(OH)2 produced:
- From the balanced equation, 2 moles of NaOH produce 1 mole of Fe(OH)2.
- Moles of Fe(OH)2: ( 0.3188 , \text{mol} \times \frac{1 , \text{mol Fe(OH)2}}{2 , \text{mol NaOH}} = 0.1594 , \text{mol} )
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Convert moles of Fe(OH)2 to grams:
- Mass of Fe(OH)2: ( 0.1594 , \text{mol} \times 89.847 , \text{g/mol} = 14.31 , \text{g} )
Therefore, approximately 14.31 grams of Fe(OH)2 was produced.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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