Multiple Choice
How many grams of water are produced when 0.23 moles of O2 react with excess H2 according to the balanced equation: 2H2 + O2 → 2H2O?
3. Chemical Reactions
Stoichiometry
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How many milliliters of 0.200 M FeCl3 are needed to react with an excess of Na2S to produce 1.38 g of Fe2S3 if the percent yield for the reaction is 65.0%? 3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)
A
50.0 mL
B
75.0 mL
C
125.0 mL
D
100.0 mL
Verified step by step guidance1
First, calculate the moles of Fe2S3 produced using its molar mass. The molar mass of Fe2S3 can be found by adding the molar masses of 2 Fe atoms and 3 S atoms. Use the formula: \( \text{moles of Fe2S3} = \frac{\text{mass of Fe2S3}}{\text{molar mass of Fe2S3}} \).
Next, account for the percent yield. Since the percent yield is 65.0%, the actual moles of Fe2S3 produced is less than the theoretical moles. Use the formula: \( \text{actual moles} = \frac{\text{moles of Fe2S3}}{\text{percent yield}} \).
Using the stoichiometry of the balanced chemical equation, determine the moles of FeCl3 required. The equation shows that 2 moles of FeCl3 produce 1 mole of Fe2S3. Therefore, use the ratio: \( \text{moles of FeCl3} = 2 \times \text{actual moles of Fe2S3} \).
Calculate the volume of FeCl3 solution needed using its molarity. The molarity equation is \( M = \frac{\text{moles}}{\text{volume in liters}} \). Rearrange to find volume: \( \text{volume} = \frac{\text{moles of FeCl3}}{\text{molarity}} \).
Convert the volume from liters to milliliters by multiplying by 1000, since 1 liter = 1000 milliliters.
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