There are several ways to do this and this method I'm showing you may be the longest but it's the first one I came up with. So here goes. This is a two equation problem and you solve the two equations simultaneously. Here are the steps.
1. Set up equations to calculate g CaC2O4 and g MgC2O4.
2. Then convert g CaC2O4 to g CaO
3. Convert g MgC2O4 to g MgO
4. Add g CaO + g MgO to find total. The correct answer is in the choices on your list.
Here are the equations.
Let X = mass CaC2O4
and Y = mass MgC2O4, then
-------------------
eqn 1 is X + Y = 0.60g
For eqn 2 you want to chemically convert X (which is CaC2O4) to CaCO3 and Y(which is MgC2O4) to MgCO3 because you know what the carbonates weigh.
MM stands for molar mass.
(MM CaCO3/MM CaC2O4)X + (MM MgCO3/MM MgC2O4)Y = 0.465g
Solve those two equations to find X and Y.
Then Xg CaCO3 can be converted to g CaO by (MM CaO/MM CaCO3)X = ?
Yg MgCO3 can be converted to g MgO by
(MM MgO/MM MgCO3)Y = ?
Then add g CaO to g MgO to find total.
You should obtain an answer VERY close to one of the choices.
Post your work if you get stuck.
A 0.60 g sample consisting of only CaC2O4 & MgC2O4 is heated at 5000C, converting the two salts of CaCO3 and MgCO3 . The sample then weighs 0.465 g. If the sample had been heated to 9000C, where the products are CaO & MgO, what would the mixtures of oxides have weighed?
a) 0.12 g b) 0.21 g c) 0.252 g d) 0.3 g
5 answers
0.3
0.3
0.3
0.252