\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ n_{H_2SO_4} = n_{ZnSO_4} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ V_{dd\ H_2SO_4}= \dfrac{0,1}{0,25} = 0,4(lít)\\ b) m_{ZnSO_4} = 0,1.161 = 16,1(gam)\)
\(n_{ZnSO_4}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(0.1.......0.1..............0.1\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.1}{0.25}=0.4\left(l\right)\)
\(m_{ZnSO_4}=0.1\cdot161=16.1\left(g\right)\)
a. Ta có nZn=6,5/65=0.1 mol
PTHH: Zn+H2SO4--->ZnSO4+H2
Theo pthh trên, nZn=nH2SO4=0,1 mol
=> VH2SO4=0.1/0.25=0,4 lít
b. nZn=nZnSO4=0.1 mol
=> mZnSO4=0.1*161=16.1 (g)
