\(\begin{array}{l}
1)\\
BaO + {H_2}O \to Ba{(OH)_2}\\
nBa{(OH)_2} = nBaO = 2\,mol\\
mBa{(OH)_2} = 2 \times 171 = 342g\\
ABa{(OH)_2} = 2 \times 6 \times {10^{23}} = 1,2 \times {10^{24}}\\
2)\\
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}\\
n{P_2}{O_5} = \frac{{71}}{{142}} = 0,5\,mol\\
= > n{H_3}P{O_4} = 1\,mol\\
m{H_3}P{O_4} = 1 \times 98 = 98g\\
3)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
n{H_2} = \frac{{3,36}}{{22,4}} = 0,15\,mol\\
= > nAl = 0,1\,mol\\
mAl = 0,1 \times 27 = 2,7g
\end{array}\)