Đáp án:
Giải thích các bước giải:
Bài `2`
`a.M_{Cu(NO_3)_2}=64+2.(14+16.3)=188(g`$/$`mol)`
`b.M_{MgSO_4}=24+12+16.3=120(g`$/$`mol)`
`c.M=28/{0,5}=n/m=56(g`$/$`mol)`
`d.M=11/{0,25}=n/m=44(g`$/$`mol)`
Bài `3`
`a.m_{Fe_2(SO_4)_3}=n.M=0,1.400=40(g)`
`b.`
`n_{NH_3}=V/{22,4}=\frac{13,44}{22,4}=0,6(mol)`
`m_{NH_3}=n.M=0,6.17=10,2(g)`
`c.`
`896(ml)=8,96(l)`
`n_{O_2}=V/{22,4}=\frac{8,96}{22,4}=0,4(mol)`
`m_{O_2}=n.M=0,4.32=12,8(g)`
`d.`
`n_{ZnCl_2}=\frac{N}{6.10^23}=\frac{8.10^23}{6.10^23}=4/3(mol)`
`m_{ZnCl_2}=n.M=4/3.136=181,333(g)`
`#Devil`
