Câu 1)
\(4Na + {O_2}\xrightarrow{{{t^o}}}2N{a_2}O\)
\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)
\(2NaOH + C{O_2}\xrightarrow{{}}N{a_2}C{O_3} + {H_2}O\)
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
\(N{a_2}O + 2HCl\xrightarrow{{}}2NaCl + {H_2}O\)
Câu 2:
Phản ứng xảy ra:
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Ta có:
\({n_{{H_2}S{O_4}}} = 0,05.1 = 0,05{\text{ mol}} \to {{\text{n}}_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,1{\text{ mol}} \to {{\text{m}}_{NaOH}} = 0,1.40 = 4{\text{ gam}} \to {m_{dd}}_{NaOH} = \frac{4}{{16\% }} = 25{\text{ gam}}\)
Nếu thay NaOH bằng KOH
\(2KOH + {H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + 2{H_2}O\)
Ta có:
\({n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,1{\text{ mol}} \to {{\text{m}}_{KOH}} = 0,1.56 = 5,6{\text{ gam}} \to {{\text{m}}_{dd{\text{KOH}}}} = \frac{{5,6}}{{5,6\% }} = 100{\text{ gam}} \to {{\text{V}}_{dd{\text{ KOH}}}} = \frac{{100}}{{1,045}} = 95,7{\text{ ml}}\)
Câu 3:
Gọi kim loại là R, suy ra oxit là RO.
Phản ứng xảy ra:
\(RO + 2HCl\xrightarrow{{}}RC{l_2} + {H_2}O\)
Ta có:
\({m_{HCl}} = 50.14,6\% = 7,3{\text{ gam}} \to {{\text{n}}_{HCl}} = \frac{{7,3}}{{36,5}} = 0,2{\text{ mol}} \to {{\text{n}}_{RO}} = \frac{1}{2}{n_{HCl}} = 0,1{\text{ mol}} \to {{\text{M}}_{RO}} = R + 16 = \frac{{7,2}}{{0,1}} = 72 \to R = 56 \to Fe\)
Vậy oxit là FeO.
