0.70 g sample consisting of `CaC_2O_4` and `MgC_2O_4` is heated at 300 °C to convert the salts to `CaCO_3` and `MgCO_3`, respectively. The sample then weighs 0.47 g. When the sample is heated to 700 °C, then the products are CaO and MgO, respectively. What is the weight of mixture of the oxides?

Options

A. 0.36 g
B. 0.14 g
C. 0.28 g
D. 1.08 g

Answer

C

Explanation

While solving this problem, students are advised to use the mole concept and stoichiometry as follows.
Consider the arbitrary mass of `MgC_(2)O_(4)` and `CaC_(2)O_(4)` to be (0.7 – x) and x, respectively.
Write the chemical reaction and calculate the weight of both the species.
Finally, using the information provided in the question complete the further calculation.
Let x g be the weight of `CaC_2O_4` present.
Then, weight of `MgC_2O_4 = (0.7 – x) g`
`CaC_(2)O_(4) overset(Delta) to CaCO_(3) + CO`
`MgC_(2)O_(4) overset(Delta) to MgCO_(3) + CO`
Weight of `CaCO_(3)` produced= `x/128 xx 100`
Weight of `MgCO_(3)` produced `= (0.7-x)/112 xx 84`
`x/128 xx 100 + (0.7-x)/112 xx 84 = 0.47`
`x = 0.46 g`
Molecular weight of `CaCO_(3) = 100 g`
Molecular weight of `MgCO_3 = 84 g`
Molecular weight of `CaC_2O_4 = 128 g`
Molecular weight of `MgC_2O_4 = 112 g`
Due to further heating,
`CaC_(2)O_(4) to CaCO_(3) overset(Delta) to CaO + CO_(2) rArr` Moles of CaO `=x/128 (x/128)`
`MgC_(2)O_(4) to MgCO_(3) overset(Delta) to MgO + CO_(2) rArr` Moles of MgO `=(0.7 -x)/112`
Weight of CaO and MgO
`=0.46/128 xx 56 + 0.24/112 xx 40`
`=0.20 + 0.0857 = 0.28 g`