Đáp án:
m=41,05 gam
Giải thích các bước giải:
\(2Mg + {O_2}\xrightarrow{{{t^o}}}2MgO\)
\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)
\(2Cu + {O_2}\xrightarrow{{{t^o}}}2CuO\)
\(MgO + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}O\)
\(A{l_2}{O_3} + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}O\)
\(CuO + 2HCl\xrightarrow{{}}2CuC{l_2} + {H_2}O\)
BTKL: \({m_{kim{\text{ loại}}}} + {m_{{O_2}}} = {m_{oxit}}\)
\( \to {m_{{O_2}}} = 17,4 - 10,52 = 6,88{\text{ gam}} \to {{\text{n}}_{{O_2}}} = 0,215{\text{ mol}}\)
Ta có: \(2{n_{{O_2}}} = {n_{MgO}} + 3{n_{A{l_2}{O_3}}} + {n_{CuO}} = 0,43{\text{ mol}}\)
Lại có \({n_{HCl}} = 2{n_{MgO}} + 6{n_{A{l_2}{O_3}}} + 2{n_{CuO}} = 0,86{\text{ mol}}\)
\( \to {n_{{H_2}O}} = \frac{1}{2}{n_{HCl}} = 0,43{\text{ mol}}\)
BTKL: \({m_{oxit}} + {m_{HCl}} = {m_{muối}} + {m_{{H_2}O}}\)
\( \to 17,4 + 0,86.36,5 = m + 0,43.18 \to m = 41,05{\text{ gam}}\)
