Zn+S--->ZnS
n S=6,4/32=0,2(mol)
n Zn=2,6/65=0,04(mol)
-->S dư
n S=n Zn=0,04(mol)
n S dư=0,2-0,04=0,16(mol)
m S dư=0,16.32=5,12(g)
\(Zn+S\underrightarrow{^{to}}ZnS\)
\(n_S=\frac{6,4}{32}=0,2\left(mol\right);n_{Zn}=\frac{2,6}{65}=0,04\left(mol\right)\)
Vì nS > nZn \(\Rightarrow\) S dư
\(\Rightarrow n_{S_{pu}}=n_{Zn}=0,04\left(mol\right)\)
\(\Rightarrow n_{S_{du}}=0,2-0,04=0,16\left(mol\right)\)
\(\Rightarrow m_{S_{du}}=0,16.32=5,12\left(g\right)\)