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@48053279
July 2022
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Какой объем газа выделится при растворении Al в 250 г. 5% растворе HCl?
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SmashMe
2Al + 6HCl = 2AlCl3 + 3H2
m с примесями(HCl)=250г
m чист(HCl)=250*0.05 = 12.5
M(HCl) = 36.5. 6 моль => 219
Vm(H2) =22,4. 3 моль =>67.2
12.5/219=x/67.2
x=3,84 л
2 votes
Thanks 1
achilles6eldor
2 AL +6 HCL ----> 2ALCL3 + 3H2 m(hcl) = 250 * 0,05 = 12,5 gr --------- 0,342
6------------------------------67,2 litr ( 3 * 22,4 )
0,342-----------------------x = 3,83 litr
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Answers & Comments
m с примесями(HCl)=250г
m чист(HCl)=250*0.05 = 12.5
M(HCl) = 36.5. 6 моль => 219
Vm(H2) =22,4. 3 моль =>67.2
12.5/219=x/67.2
x=3,84 л
6------------------------------67,2 litr ( 3 * 22,4 )
0,342-----------------------x = 3,83 litr