Question

Zn and hydrochloric acid react according to the reaction, Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H (9) If 0.30 mol Zn are added to HCl containing 0.52 mol of HCI how many moles of H, are produced?​

Answer 1

Answer:

Zn( s) + 2HCl(aq) --> ZnCl2(aq) + H2(g)

In this reaction find out the limiting reagent as follows:

0.3 mol Zn x 1mol H2/1 mole Zn = 0.3 mole H2

0.52 mole H2 x 1 mole H2/2 mole HCl = 0.26 mole H2

Explanation:

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