Question #a5115

1 Answer
Stefan V.
Mar 22, 2017

Use its molar mass!

Explanation:

Let's say that you're dealing with a generic reaction

#"A " + " 2B " -> " C"#

Notice that you have a #1:1# mole ratio between #"A"# and #"C"#. This tells you that the reaction will consume a number of moles of #"A"# and produce the same number of moles of #"C"#.

So, let's say that the reaction consumes #3# moles of #"A"#. The #1:1# mole ratio tells you that it will produce #3# moles of #"C"#.

To convert the number of moles of #"C"# to grams, use its molar mass. Let's say that #"C"# has a molar mass of #x# #"g mol"^(-1)#.

This means that #1# mole of #"C"# has a mass of #x# grams. You can thus say that the reaction produced

#3 color(red)(cancel(color(black)("moles C"))) * overbrace((x color(white)(.)"g")/(1color(red)(cancel(color(black)("mole C")))))^(color(blue)("the molar mass of C")) = (3x)" g"#

of the product, i.e. #(3x)# #"g"# of #"C"#.

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