a)
mNaCl = M*n= 2*58.5=117g
b)
nH2= 5.6/22.4=0.25 mol
a) \(m_{NaCl}=2\times58,5=117\left(g\right)\)
b) \(n_{H_2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
a) ta có : nNaCl = 2 mol
=> mNaCl = 2 . 58,5 = 117 g
b) ta có VH2 = 5,6 lít
=> nH2 = \(\frac{V_{H2}}{22,4}=\frac{5,6}{22,4}=0,25mol\)