Question

Question #719a9

Answer 1

`1.0156` grams `K_2CO_3`

Explanation:

1. Take the balanced equation;
   `2KHCO_3->K_2CO_3+H_2CO_3`
2. Take the molar masses of `KHCO_3` and `K_2CO_3`
3. Compute stoichiometrically, starting with the mass of `KHCO_3`
   `1.6cancel(g KHCO_3)`x`(1 cancel(mol KHCO_30)/(cancel(100g KHCO_3))`x`(1 cancel(mol K_2CO_3))/(2 cancel(mol KHCO_3))`x`(138g K_2CO_3)/(1 cancel(mol K_2CO_3))`
4. The answer is `1.104` g `K_2CO_3`
5. To compute the atual yield, use the formula:
   `% yield=(actual yield)/(theo yield)`, rearrange and plug in values to the derived formula;
   `actual yied=(% yield)(theo yield)`
6. The actual yield is `1.0156g` `K_2CO_3`