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?

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.