Asked by Anna Diaz
7.5 L of hydrogen gas at STP and zinc (II) nitrate was produced by reacting zinc with nitric acid. Calculate the mass of zinc needed for this reaction.
This is what I did: Zn + 2HNO3 = H2 + Zn(NO3)2.
7.5L H2 * 1 mole H2/ 22.4 L H2 * 1 mole Zn/ 1 mole H2 * 65.4 g Zn / 1 mole Zn
I got 21.9 g Zn but I am not sure it is right.
This is what I did: Zn + 2HNO3 = H2 + Zn(NO3)2.
7.5L H2 * 1 mole H2/ 22.4 L H2 * 1 mole Zn/ 1 mole H2 * 65.4 g Zn / 1 mole Zn
I got 21.9 g Zn but I am not sure it is right.
Answers
Answered by
DrBob222
That looks good to me but you have too many significant figures if you posted the problem correctly. 7.5 L allows only two places and I would round to 22 g.
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