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?

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?

Owen Ambrose · Accepted Answer

#"105 g CaCl"_2#

Write down the balanced chemical equation that best represents your reaction first.

#"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.

The two compounds' molar masses can be used to convert this mole ratio to a gram ratio.

#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"#

Because you don't have as much hydrochloric acid as you would need to react with the entire mass of calcium oxide

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

It is possible to deduce that the hydrochloric acid will function as a limiting reagent, meaning that it will be fully dissolved before the grams of calcium oxide have an opportunity 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'll possess

#(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

You could say that the reaction will produce because all of the hydrochloric acid will react.

#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)|)))#

Three sig figs are used to round the result.

Naomi Aronson · Answer

undefined To find the amount of CaCl2 produced, you need to determine the limiting reactant first. Then, use stoichiometry to calculate the amount of CaCl2 produced based on the balanced equation.

1. Calculate the moles of CaO: $ \text{moles} = \frac{\text{mass}}{\text{molar mass}} $
2. Calculate the moles of HCl: $ \text{moles} = \frac{\text{mass}}{\text{molar mass}} $
3. Determine the limiting reactant by comparing the moles of CaO and HCl.
4. Use the mole ratio from the balanced equation to find the moles of CaCl2 produced.
5. Convert moles of CaCl2 to grams: $ \text{mass} = \text{moles} \times \text{molar mass} $

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

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?