Đáp án:
\(\begin{array}{l}
2)\\
{m_{Ba{{(OH)}_2}}} = 342g\\
{A_{Ba{{(OH)}_2}}} = 1,2 \times {10^{24}}\\
3)\\
{m_{{H_3}P{O_4}}} = 98g\\
4)\\
{m_{Al}} = 2,7g\\
5)\\
{m_{FeS{O_4}}} = 3,04g\\
{V_{{H_2}}} = 0,448l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2)\\
BaO + {H_2}O \to Ba{(OH)_2}\\
{n_{Ba{{(OH)}_2}}} = {n_{BaO}} = 2\,mol\\
{m_{Ba{{(OH)}_2}}} = n \times M = 171 \times 2 = 342g\\
{A_{Ba{{(OH)}_2}}} = n \times 6 \times {10^{23}} = 2 \times 6 \times {10^{23}} = 1,2 \times {10^{24}}\\
3)\\
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}\\
{n_{{P_2}{O_5}}} = \dfrac{m}{M} = \dfrac{{71}}{{142}} = 0,5\,mol\\
{n_{{H_3}P{O_4}}} = 2{n_{{P_2}{O_5}}} = 0,5 \times 2 = 1\,mol\\
{m_{{H_3}P{O_4}}} = n \times M = 1 \times 98 = 98g\\
4)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Al}} = \dfrac{{0,15 \times 2}}{3} = 0,1\,mol\\
{m_{Al}} = n \times M = 0,1 \times 27 = 2,7g\\
5)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{1,12}}{{56}} = 0,02\,mol\\
{n_{FeS{O_4}}} = {n_{{H_2}}} = {n_{Fe}} = 0,02\,mol\\
{m_{FeS{O_4}}} = n \times M = 0,02 \times 152 = 3,04g\\
{V_{{H_2}}} = n \times 22,4 = 0,02 \times 22,4 = 0,448l
\end{array}\)
