nH2= 0,56 : 22,4 = 0,025 mol
mH2= 0,025 . 2=0,05g
nhcl=6,72:22,4=0,2 mol
mhcl=0,2 . 36,5 = 7,3g
a) \(n_{H_2}=\frac{0,56}{22,4}=0,025\left(mol\right)\)
\(\Rightarrow m_{H_2}=n.M=0,025\times2=0,05\left(gam\right)\)
b) \(n_{HCl}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow m_{HCl}=n.M=0,3\times36,5=10,95\left(gam\right)\)
a.nH2=V:22,4=0,025mol
mH2=n.M=0,025.2=0,05g
b.nHCl=V:22,4=6,72:22,4=0,3mol
mHCl=n.M=0,3.36,5=10,95g