Question #e5006

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Answer 1 · 2017-12-25T04:11:30

$=96.2gCaCO_3$

Write the skeleton and balance the given equation
$Ca(OH)_2(s)+CO_2(g)>CaCO_3(s)+H_2O(g)$
Find the molar mass of $CaCO_3$ as the involved compound as specified in the problem.
$=(100g)/(mol)$
Since the given data is already expressed in $mol$ , conversion per convention is no longer necessary in this case.
Referring to the balanced equation shown above for the mole ratio, find the $etaCaCO_3$ through molar conversion; i.e.,
$=0.962cancel(molCa(OH))_2xx(1molCaCO_3)/(1cancel(molCa(OH)_2))$
$=0.962molCaCO_3$
Now, find the $"mass"(m)$ of $CaCO_3$ , knowing the $etaCaCO_3$ as computed, and the molar mass of the compound for the conversion.
$=0.962cancel(molCaCO_3)xx(100gCaCO_3)/(1cancel(molCaCO_3))$
$=96.2gCaCO_3$