1 mol chất chứa \(6,023.10^{23}\) nguyên tử
a)
\({n_{Fe}} = \frac{{{{5,2.10}^{23}}}}{{{{6,023.10}^{23}}}} = 0,86336{\text{ mol}}\)
b)
\({n_{NaOH}} = \frac{{{{12,4.10}^{23}}}}{{{{6,023.10}^{23}}}} = 2,0588{\text{ mol}}\)
c)
\({n_{Al}} = \frac{{{m_{Al}}}}{{{M_{Al}}}} = \frac{{5,4}}{{27}} = 0,2{\text{ mol}}\)
d)
\({n_{CaS{O_4}}} = \frac{{{m_{CaS{o_4}}}}}{{{n_{CaS{O_4}}}}} = \frac{{23,8}}{{40 + 32 + 16.4}} = 0,175{\text{ mol}}\)
e)
\({n_{{O_2}}} = \frac{{{V_{{O_2}}}}}{{22,4}} = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol}}\)
f)
\({n_{C{l_2}}} = \frac{{7,2}}{{22,4}} \to {m_{C{l_2}}} = \frac{{7,2}}{{22,4}}.71 = 22,82{\text{ gam}}\)
g)
\({n_{HCl}} = {V_{dd}}.{C_{M{\text{ HCl}}}} = 0,2.0,8 = 0,16{\text{ mol}}\)
h)
\({m_{NaOH}} = {m_{dd{\text{NaOH}}}}.C{\% _{NaOH}} = 200.10\% = 20{\text{ gam}}\)
\( \to {n_{NaOH}} = \frac{{{m_{NaOH}}}}{{{M_{NaOH}}}} = \frac{{20}}{{23 + 16 + 1}} = 0,5{\text{ mol}}\)
i)
\({m_{dd\;{\text{NaOH}}}} = 300.1,1 = 330{\text{ gam}}\)
\({m_{NaOH}} = {m_{dd{\text{NaOH}}}}.C{\% _{NaOH}} = 330.8\% = 26,4{\text{ gam}}\)
\( \to {n_{NaOH}} = \frac{{{m_{NaOH}}}}{{{M_{NaOH}}}} = \frac{{26,4}}{{23 + 16 + 1}} = 0,66 {\text{ mol}}\)
