An Unimaginatively named Chemistry Problem- Mole Concept.

This problem is a cakewalk using the POAC method. There is another way of doing it which is quite algebraic and borders on stupidity.

Let there be x gram of $Ca{ Cl }_{ 2 }$ in the original mixture

We work our way backwards over here, so using POAC :

$Ca{ CO }_{ 3 }\longrightarrow CaO$

Balancing Ca :

Suppose y gram of $Ca{ CO }_{ 3 }$ produced 1.62 gram of CaO $1\times \frac { y }{ 100 } =1\times \frac { 1.62 }{ 56 } \\ \Rightarrow y=\frac { 162 }{ 56 }$

Now we know how much of $Ca{ CO }_{ 3 }$ was produced by the reaction between $Ca{ Cl }_{ 2 }$ and ${ Na }_{ 2 }{ CO }_{ 3 }$

So again using POAC

$Ca{ Cl }_{ 2 }\longrightarrow Ca{ CO }_{ 3 }$

Balancing Ca again,

$1\times \frac { x }{ 111 } =1\times \frac { 162 }{ 5600 } \\ \Rightarrow x\approx 3.21$

Therefore % of $Ca{ Cl }_{ 2 }$ in the original mixture=

$\frac { 3.21 }{ 10 } \times 100$ % = 32.1%

Using the molar mass of $\ce{CaO}$ as $56$ g/mol, then $1.62$ g of $\ce{CaO} = \dfrac{1.62}{56} = 0.0289$ mol.

Therefore, there is $0.0289$ mol of $\ce{CaCl2}$ (molar mass $= 111$ g/mol) or $0.0289 \times 111 = 3.2111$ g in the original sample, which is $\dfrac{3.2111}{10} \approx \boxed{32.1}$ %.

Feed

((((1.62 g) / (CaO molar mass)) * (CaCl2 molar mass)) / (10 g)) * 100

To Wolfram|Alpha.

Step-by-step already written by Siddharth.