Question #a2839

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Answer 1 · 2016-10-08T19:06:57

$0.71 g$ $0.71"g"$

From the answer to the previous question, we know the moles of sulfuric acid = $0.705$ $0.705$ .

$2 Al + 3 H_{2} {SO}_{4} \to {Al}_{2} {({SO}_{4})}_{3} + 3 H_{2}$ $2"Al" + 3"H"_2"SO"_4 rarr "Al"_2("SO"_4)_3 + 3"H"_2$

Using this, we can work out the moles of hydrogen:

Reacting ratio of $"H"_2"SO"_4: "H"_2 = 3:3 = 1:1 therefore$

moles of $"H"_2 = 0.705$

$"n"="m"/"M"_"r" rarr 0.705 = "m"/1$

$"m"=0.705*1=0.705"g"$