Question #7fb69
1 Answer
Explanation:
The idea here is that you need to use the pecent composition of copper in copper(I) oxide,
So, use the molar mass of copper and those of copper(I) oxide and copper(II) oxide, respectively, to find the percent compositions of copper in the two oxides
#"For Cu"_2"O: " ( 2 xx 63.55color(red)(cancel(color(black)("g/mol"))))/(143.1color(red)(cancel(color(black)("g/mol")))) xx 100 = "88.8% Cu"#
#"For CuO: " (63.55color(red)(cancel(color(black)("g/mol"))))/(79.55color(red)(cancel(color(black)("g/mol")))) xx 100 = "79.9% Cu"#
So, let's say that
#x + y = "9.80 g"#
Now, use the percent compositions of copper to write a second equation using the mass of copper you know the mixture contains
#0.888 * x + 0.799 * y = "8.22 g"#
This system of equations can be solved by substitution
#x = 9.80 - y#
#0.888 * (9.80 - y) + 0.799 * y = 8.22#
#8.702 - 0.888 * y + 0.799 * y = 8.22#
#0.089 * y = 0.482 implies y = "5.4157 g"#
The mass of copper(II) oxide in the mixture is thus equal to - rounded to three sig figs
#m_"CuO" = color(green)("5.42 g")#