1)
Phản ứng xảy ra:
\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)
\(2Cu + {O_2}\xrightarrow{{{t^o}}}2CuO\)
Ta có:
\({n_{Fe}} = \frac{x}{{56}} \to {n_{Fe{\text{ phản ứng}}}} = \frac{x}{{56}}.80\% = \frac{x}{{70}}{\text{ mol}}\)
\({n_{Cu}} = \frac{x}{{64}} \to {n_{Cu{\text{ phản ứng}}}} = \frac{x}{{64}}.80\% = \frac{x}{{80}}{\text{ mol}}\)
\( \to {n_{F{e_3}{O_4}}} = \frac{1}{3}{n_{Fe{\text{ phản ứng}}}} = \frac{1}{3}.\frac{x}{{70}} = \frac{x}{{210}}{\text{ mol}}\)
\( \to {m_{F{e_3}{O_4}}} = \frac{x}{{210}}.(56.3 + 16.4) = \frac{{116x}}{{105}}\)
\( \to {n_{CuO}} = {n_{Cu}} = \frac{x}{{80}} \to {m_{CuO}} = \frac{x}{{80}}.80 = x{\text{ gam < }}{{\text{m}}_{F{e_3}{O_4}}}\)
Sản phẩm của \(Fe\) có khối lượng lớn hơn.
2)
Phản ứng xảy ra:
\(2Mg + {O_2}\xrightarrow{{{t^o}}}2MgO\)
\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)
Ta có:
\({n_{MgO{\text{ lt}}}} = {n_{Mg}} = a{\text{ mol}}\)
\( \to {n_{MgO}} = a.80\% = 0,8a{\text{ }}mol\)
\( \to {m_{MgO}} = 0,8a.40 = 32a{\text{ gam}}\)
\({n_{A{l_2}{O_3}{\text{ lt}}}} = \frac{1}{2}{n_{Al}} = 0,5a{\text{ mol}}\)
\( \to {n_{A{l_2}{O_3}}} = 0,5a.65\% = 0,325a{\text{ mol}}\)
\( \to {m_{A{l_2}{O_3}}} = 0,325a.(27.2 + 16.3) = 33,15a{\text{ gam}}\)
