\(2Na+2HCl-->2NaCl+H2\)
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{Na}=2n_{H2}=0,6\left(mol\right)\)
\(m_{Na}=0,6.23=13,8\left(g\right)\)
2Na+2HCl−−>2NaCl+H22Na+2HCl−−>2NaCl+H2
nH2=6,7222,4=0,3(mol)nH2=6,7222,4=0,3(mol)
nNa=2nH2=0,6(mol)nNa=2nH2=0,6(mol)
mNa=0,6.23=13,8(g)
Mk sao chép đấy kkk:)))