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What is the mass of iron that took part in the reaction and show that the percentage of...
What is the mass of iron that took part in the reaction and show that the percentage of iron in the alloy is 88.1% in the following case:
Silicon steel is an alloy of the elements iron, carbon and silicon. An alloy sample was reacted with excess hydrochloric acid and the following reaction occurred:
Fe (s) + 2HCl (aq) ---> FeCl2 (aq) + H2 (g)
The carbon and silicon in the alloy did not react with the acid. If an alloy sample with a mass of 0.160 g produced 62.0 mL of hydrogen gas, measured at SLC, calculate the mass of iron that reacted to produce this amount of hydrogen where the amount of hydrogen evolved in the reaction is 2.53 x 10^-3 mol as well as the percentage of iron in the alloy.
I’ve asked this question yesterday but wanted to re-ask it because when we use the PV = nRT formula, I thought R = 8.31 rather than 82.05746, can you also please tell me why this is so? Also when the question was answered for the percentage of the iron in the alloy, it was answered 44.275% where the actual answer was 88.1%, unless the textbook has made an error itself.
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High School Teacher
You know that 2.53 x 10^-3 moles of hydrogen are produced. Looking at the balanced chemical equation, for every mole of H2 produced, one mole of Fe is consumed. A 1:1 ratio.
Therefore, you started with 2.53 x 10^-3 moles of Fe (assuming 100% reaction between the Fe and HCl).
2.53 x 10^-3 moles of Fe * 55.847 g/mole of Fe = 0.141 g of Fe in the alloy.
Since the total mass of the alloy is 0.16 g, that means the % Fe is:
0.141/0.16 * 100 = 88.3%, the difference between this and the book answer due to rounding.
Posted by ndnordic on January 10, 2012 at 5:50 AM (Answer #1)
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