Question

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?

Answer 1

`"105 g CaCl"_2`

Explanation:

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

`"CaO"_ ((s)) + color(red)(2)"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l))`

You need `color(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:color(red)(2)` mole ratio will thus be equivalent to

`(56.08 color(red)(cancel(color(black)("g mol"^(-1)))))/(color(red)(2) * 36.46color(red)(cancel(color(black)("g mol"^(-1))))) = 56.08/72.92 ->` gram ratio

So, you need `"72.92 g"` of hydrochloric acid for every `"56.08 g"` of calcium oxide that take part in the reaction.

You know that you have a sample of `"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 `color(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

`(color(red)(2) * 36.46 color(red)(cancel(color(black)("g mol"^(-1)))))/(111.0color(red)(cancel(color(black)("g mol"^(-1))))) = 72.92/111.0 ->` 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.