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Question #c8da8

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Answer 1 · 2017-01-13T20:37:24

The balanced equation is:

$C a {(O H)}_{2} + C O_{2} \to C a C O_{3} + H_{2} O$ $Ca(OH)_2 + CO_2 to CaCO_3 + H_2O$

Explanation:

From the equation, please observe that 1 mole of $C a {(O H)}_{2}$ $Ca(OH)_2$ produces:

$\frac{0.962 mole C a {(O H)}_{2}}{1} \frac{1 mole C a C O_{3}}{1 mole C a {(O H)}_{2}}$ $(0.962" mole "Ca(OH)_2)/1(1" mole "CaCO_3)/(1" mole "Ca(OH)_2)$

The molar mass of $C a C O_{3}$ $CaCO_3$ is 100 g:

$(0.962cancel(" mole "Ca(OH)_2))/1(1cancel(" mole "CaCO_3))/(1cancel(" mole "Ca(OH)_2))(100" g "CaCO_3)/(1cancel(" mole "CaCO_3)) = 96.2" g "CaCO_3$

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