+ 26 gam \(BaCl_2\)
Ta có:
\({n_{BaC{l_2}}} = \frac{{26}}{{137 + 35,5.2}} = 0,125{\text{ mol}}\)
\( \to {n_{Ba}} = {n_{BaC{l_2}}} = 0,125{\text{ mol;}}{{\text{n}}_{Cl}} = 2{n_{BaC{l_2}}} = 0,125.2 = 0,25{\text{ mol}}\)
\( \to {m_{Ba}} = 0,125.137 = 17,125{\text{ gam;}}{{\text{m}}_{Cl}} = 0,25.35,5 = 8,875{\text{ gam}}\)
+ 8 gam \(Fe_2O_3\)
Ta có:
\({n_{F{e_2}{O_3}}} = \frac{8}{{56.2 + 16.3}} = 0,05{\text{ mol}}\)
\( \to {n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,1{\text{ mol;}}{{\text{n}}_O} = 3{n_{F{e_2}{O_3}}} = 0,15{\text{ mol}}\)
\( \to {m_{Fe}} = 0,1.56 = 5,6{\text{ gam;}}{{\text{m}}_O} = 0,15.16 = 2,4{\text{ gam}}\)
+ 4,4 gam \(CO_2\)
\( \to {n_{C{O_2}}} = \frac{{4,4}}{{12 + 16.2}} = 0,1{\text{ mol}}\)
\( \to {n_C} = {n_{C{O_2}}} = 0,1{\text{ mol;}}{{\text{n}}_O} = 2{n_{C{O_2}}} = 0,2{\text{ mol}}\)
\( \to {m_C} = 0,1.12 = 1,2{\text{ gam;}}{{\text{m}}_O} = 0,2.16 = 3,2{\text{ gam}}\)
+ 7,56 gam \(MnCl_2\)
Ta có:
\({n_{MnC{l_2}}} = \frac{{7,56}}{{55 + 35,5.2}} = 0,06{\text{ mol}}\)
\( \to {n_{Mn}} = {n_{MnC{l_2}}} = 0,06{\text{ mol;}}{{\text{n}}_{Cl}} = 2{n_{MnC{l_2}}} = 0,12{\text{ mol}}\)
\( \to {m_{Mn}} = 0,06.55 = 3,3{\text{ gam}}\)
\({m_{Cl}} = 0,06.35,5 = 2,13{\text{ gam}}\)
+5,6 gam \(NO
\( \to {n_{NO}} = \frac{{5,6}}{{14 + 16}} = \frac{{5,6}}{{30}}\)
\( \to {n_N} = {n_O} = {n_{NO}} = \frac{{5,6}}{{30}}{\text{ mol}}\)
\( \to {m_N} = \frac{{5,6}}{{30}}.14 = 2,6133{\text{ gam;}}{{\text{m}}_O} = 5,6 - 2,6133 = 2,9867{\text{ gam}}\)
