\(n_{Zn}=\dfrac{48,75}{65}=0,75mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,75 0,75 ( mol )
\(V_{H_2}=0,75.22,4=16,8l\)
\(Fe_3O_4+4H_2\rightarrow\left(t^o\right)3Fe+4H_2O\)
0,75 0,5625 ( mol )
\(m_{Fe}=0,5625.56=31,5g\)
Zn+2HCl->ZnCl2+H2
0,75-------------------0,75
Fe3O4+4H2-to>3Fe+4H2O
0,75--------0,5625g
n Zn=\(\dfrac{48,75}{65}\)=0,75 mol
=>VH2=0,75.22,4=16,8l
=>m Fe=0,5625.56=31,5g