Ta có:
({n_{O(X)}} = frac{{43,56.0,2204}}{{16}} = 0,6(mol))
(Zleft{ begin{array}{l}
{N_2}O,C{O_2}:a\
NO:b
end{array} right. to left{ begin{array}{l}
a + b = 0,12\
44a + 30b = 0,12.2.18,5
end{array} right. to left{ begin{array}{l}
a = 0,06\
b = 0,06
end{array} right.)
Quy đổi hỗn hợp X, ta có sơ đồ phản ứng sau:
(Xleft{ begin{array}{l}
Mg,Fe,Cu\
O\
C{O_2}:0,06 – x
end{array} right. to Yleft{ begin{array}{l}
K{L^{m + }}\
NH_4^ + \
NO_3^ – :1,77
end{array} right. + Zleft{ begin{array}{l}
{N_2}O:x\
C{O_2}:0,06 – x\
NO:0,06
end{array} right.)
(begin{array}{l}
to {n_O} = 0,6 – 2.left( {0,06 – x} right) = 2x + 0,48\
to {n_{NH_4^ + }} = 1,92 – 1,77 – 2x – 0,06 = 0,09 – 2x
end{array})
({n_{HN{O_3}}} = 10.left( {0,09 – 2x} right) + 10x + 4.0,06 + 2.left( {2x + 0,48} right) = 1,92 to x = 0,03)
( to {n_{C{O_2}}} = 0,06 – 0,03 = 0,03)
(left{ begin{array}{l}
{m_x} = 24y + 80z + 232t + 116.0,03 = 43,56\
{m_{oxit}} = 40y + 80z + 160.left( {1,5t + 0,015} right) = 48\
{n_{O(X)}} = z + 4t + 3.0,03 = 0,6
end{array} right. to left{ begin{array}{l}
y = 0,3\
z = 0,15\
t = 0,09
end{array} right.)
( to % {m_{CuO}} = frac{{80.0,15}}{{43,56}}.100% = 27,55% )