# If I combine 15.0 grams of calcium hydroxide with 75.0 mL of 0.500M HCl, how many grams of calcium chloride would be formed?

May 30, 2018

Approx...$2 \cdot g$...

#### Explanation:

$C a {\left(O H\right)}_{2} \left(s\right) + 2 H C l \left(a q\right) \rightarrow C a C {l}_{2} \left(a q\right) + 2 {H}_{2} O$

${n}_{C a {\left(O H\right)}_{2}} = \frac{15.0 \cdot g}{74.09 \cdot g \cdot m o {l}^{-} 1} = 0.203 \cdot m o l$

And we got with respect to $H C l$ a molar quantity of...

$75.0 \cdot m L \times 0.500 \cdot m o l \cdot {L}^{-} 1 \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 = 0.0375 \cdot m o l$

Clearly the calcium hydroxide is present in molar excess....and at most we can form $\frac{0.0375 \cdot m o l}{2}$ of calcium chloride...a mass of 110.98*g*mol^-1xx(0.0375*mol)/2=??*g..

The question operates under a faulty premise...because the species in solution after addition of the acid would be $C a \left(O H\right) C l$