Question #719a9

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Answer 1 · 2016-03-16T02:06:02

$1.0156$ grams $K_2CO_3$

Take the balanced equation;
$2KHCO_3->K_2CO_3+H_2CO_3$
Take the molar masses of $KHCO_3$ and $K_2CO_3$
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 $(138g K_2CO_3)/(1 cancel(mol K_2CO_3))$
The answer is $1.104$ g $K_2CO_3$
To compute the atual yield, use the formula:
$% yield=(actual yield)/(theo yield)$ , rearrange and plug in values to the derived formula;
$actual yied=(% yield)(theo yield)$
The actual yield is $1.0156g$ $K_2CO_3$