Zn(s) + 2 HCl(aq) —> ZnCl2(aq)
+ H2(g)
If a scientist wanted to create 10L of hydrogen gas at RTP, what mass of HCl would they need to start with?

Answers

Answer 1

Answer:

First write the balanced equation for the reaction of HCl with Zn (NOTE: the previous answer did NOT do this and thus obtained an incorrect answer)

Zn + 2HCl ==> ZnCl2 + H2(g)

moles of H2 desired:  PV = nRT

n = PV/RT = (1 atm)(10L)/(0.0821 Latm/Kmol)(298K)

n = 0.409 mole H2 desired

moles Zn needed:

0.409 moles H2 x 1 mole Zn/2 moles HCl = 0.204 moles Zn needed

mass Zn needed = 0.204 moles Zn x 65.4 g/mol = 13.4 g Zn needed (to 3 sig. figs.)

Explanation:people always get this wrong but i got you


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Answers

Answer:

Melting and Boiling Point Determination

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Answers

Answer:

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Answers

Answer:

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