\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Pt : \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2|\)
1 1 1 1
0,2 0,2
\(n_{Zn}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(m_{Zn}=0,1.65=6,5\left(g\right)\)
\(m_{Ag}=15-6,5=8,5\left(g\right)\)
0/0Zn = \(\dfrac{6,5.100}{15}=43,33\)0/0
0/0Ag = \(\dfrac{8,5.100}{15}=56,67\)0/0
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