# Lets say we are given this balanced equation: CaO + 2HCl -> CaCl_2 + H_2O. 60.4g of CaO is reacted with 69.0g of HCl. How much CaCl_2 is produced?

Jun 30, 2016

${\text{105 g CaCl}}_{2}$

#### Explanation:

Start by writing out the balanced chemical equation that describes your reaction

${\text{CaO"_ ((s)) + color(red)(2)"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O}}_{\left(l\right)}$

You need $\textcolor{red}{2}$ moles of hydrochloric acid for every mole of calcium oxide that takes part in the reaction.

You an convert this mole ratio to a gram ratio by using the molar masses of the two compounds

M_("M CaO") = "56.08 g mol"^(-1)

M_("M HCl") = "36.46 g mol"^(-1)

A $1 : \textcolor{red}{2}$ mole ratio will thus be equivalent to

$\left(56.08 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g mol"^(-1)))))/(color(red)(2) * 36.46color(red)(cancel(color(black)("g mol}}^{- 1}}}}\right) = \frac{56.08}{72.92} \to$ gram ratio

So, you need $\text{72.92 g}$ of hydrochloric acid for every $\text{56.08 g}$ of calcium oxide that take part in the reaction.

You know that you have a sample of $\text{60.4 g}$ of calcium oxide available. This many grams of calcium oxide would require

60.4 color(red)(cancel(color(black)("g CaO"))) * "72.92 g HCl"/(56.08color(red)(cancel(color(black)("g CaO")))) = "78.54 g HCl"

Since you have less hydrochloric acid than you would need in order for all the mass of calcium oxide to react

overbrace("69.0 g")^(color(blue)("what you have")) < overbrace("78.54 g")^(color(darkgreen)("what you need"))

you can conclude that hydrochloric acid will act as a limiting reagent, i.e. it will be completely consumed before all the grams of calcium oxide get a chance to react.

Use the molar mass of calcium chloride to convert the $\textcolor{red}{2} : 1$ mole ratio that exists between hydrochloric acid and calcium chloride into a gram ratio

M_("M CaCl"_2) = "111.0 g mol"^(-1)

You will have

$\left(\textcolor{red}{2} \cdot 36.46 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g mol"^(-1)))))/(111.0color(red)(cancel(color(black)("g mol}}^{- 1}}}}\right) = \frac{72.92}{111.0} \to$ gram ratio

Since all the hydrochloric acid will react, you can say that the reaction will produce

69.0 color(red)(cancel(color(black)("g HCl"))) * "111.0 g CaCl"_2/(72.92color(red)(cancel(color(black)("g HCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("105 g CaCl"_2)color(white)(a/a)|)))

The answer is rounded to three sig figs.