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SteenHD
@SteenHD
July 2022
1
7
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Помогите решить по химии:
Al+H2SO4
Дано:
m(Al)с прим.=5г.
Wпр.=6%
Найти: V(H2)-?
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klestatyana
2Al+3H2SO4=Al2(SO4)3+3H2;
m(Al)чист.= 5* (1 - 0,06)=4,7 г;
M(Al)=27 г/моль;
n(Al)=m(Al) / M(Al)=4,7 / 27=0,17 моль;
n(H2)=n(Al) / 2 * 3=0,17 / 2 * 3=0,255 моль;
V(H2)=n(H2) * Vm=0,255 * 22,4=5,7 л.
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Answers & Comments
m(Al)чист.= 5* (1 - 0,06)=4,7 г;
M(Al)=27 г/моль;
n(Al)=m(Al) / M(Al)=4,7 / 27=0,17 моль;
n(H2)=n(Al) / 2 * 3=0,17 / 2 * 3=0,255 моль;
V(H2)=n(H2) * Vm=0,255 * 22,4=5,7 л.