2AL #(s)# + 6HCL#3(aq)# #rarr# 2AlCl#3(aq)# + 3H#2(g)# 0.040 mol piece of aluminium reacted completely in 20s. the rate of formation of hydrogen gas is?

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2AL #(s)# + 6HCL#3(aq)# #rarr# 2AlCl#3(aq)# + 3H#2(g)#

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Answer 1 · 2018-06-06T07:48:50

#0.003 "mol/s"#

From the equation, we can see that the ratio of number of moles:

Al : HCl : AlCl3 : H2 (Assuming that HCl3 was a typo)

is equal to:

2 : 6 : 2 : 3 (As the numbers in front of each substances show)

Therefore, if 0.040 mol of aluminium reacted completely, than #0.040 \times 3/2 = 0.060 "mol"# of hydrogen was formed.

Thus, #0.060 \divide 20 = 0.003 "mol/s"#

Hope that makes sense!