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Alina00Malina
@Alina00Malina
July 2022
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Определите массу полученного аммиака (NH3), из 50г азота (N2), содержащего 5% примесей.
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Olya880
N2+3H2--t°, p-->2NH3
m(примесей)=50*0.05=2.5 г
m(N2)=50-2.5=47.5 г
n(N2)=m/M=47.5/28≈1.7 моль
n(NH3)=1.7*2=3.4 моль
m(NH3)=n*M=3.4*17=57.8 г
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Answers & Comments
m(примесей)=50*0.05=2.5 г
m(N2)=50-2.5=47.5 г
n(N2)=m/M=47.5/28≈1.7 моль
n(NH3)=1.7*2=3.4 моль
m(NH3)=n*M=3.4*17=57.8 г