Đáp án:
\(a.\ m_{ZnCl_2}=13,6\text{(gam)}\\ b. \ V_{H_2}=2,24\text{(lít)}\\ c.\ m_{Fe}=3,73\text{(gam)}\)
Giải thích các bước giải:
\(a. \ Zn+2HCl\to ZnCl_2+H_2\\ n_{Zn}=\dfrac{6,5}{65}=0,1\text{(mol)}\to n_{ZnCl_2}=n_{Zn}=0,1\text{(mol)}\to m_{ZnCl_2}=0,1\times136=13,6\text{(gam)}\\b.\ n_{H_2}=n_{Zn}=0,1\text{(mol)}\to V_{H_2}=22,4\times 0,1=2,24\text{(lít)}\\ c.\ 3H_2+Fe_2O_3\xrightarrow{t^{\circ}}2Fe+3H_2O\\ n_{Fe}=\dfrac 23n_{H_2}=\dfrac 23\times 0,1=\dfrac 1{15}\text{(mol)}\to m_{Fe}=56\times \dfrac 1{15}=3,73\text{(gam)}\)