a) \(n_{NaCl}=\dfrac{5,85}{23+35,5}=0,1\left(mol\right)\)
b) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) \(\Rightarrow m_{H_2}=0,2\cdot2=0,4\left(g\right)\)
c) \(n_{CO_2}=\dfrac{22}{12+16\cdot2}=0,5\left(mol\right)\) \(\Rightarrow m_{CO_2}=0,5\cdot22,4=11,2\left(l\right)\)
a) nCO2=V/22,4=5,6/22,4=0,25 (mol)
b) nN2=V/22,4=2,8/22,4=0,125 (mol)
c) nO2=V/22,4=1344/22,4=60 (mol)
a) nNaCl = mNaCl : MNaCl
= 5.85 : 58.5
= 0.1 (g)