When #"11.6 g"# of citric acid reacts with excess sodium bicarbonate, what mass of #"CO"_2# is made in #"g"#? #MW = "44 g/mol"#
1 Answer
You would make
You should look for the keywords that identify what kind of problem this is.
Since it is mentioned that sodium bicarbonate is in excess, it means that the other reactant, citric acid, is the limiting reactant. That means it will be completely used up first, and that limits the amount of product you can make (
Your reaction was:
#3"NaHCO"_3(s) + "C"_6"H"_8"O"_7(aq) -> 3"CO"_2(g) + 3"H"_2"O"(l) + "Na"_3"C"_6"H"_5"O"_7(aq)#
The remainder of the question is asking you to start with a known mass of
Note that the limiting reactant must be used when determining the theoretical yield.
Each coefficient can be considered the mols of the substance involved in the reaction, so for example...
#"3 mols NaHCO"_3(s)# react with#"1 mol C"_6"H"_8"O"_7(aq)# .#"3 mols NaHCO"_3(s)# can produce#"3 mols CO"_2(g)# .#"3 mols NaHCO"_3(s)# can produce#"3 mols H"_2"O"(l)# .
and so on. This allows you to write the following unit conversion (the same units cancel out) using the molar mass of citric acid:
#11.6 cancel("g C"_6"H"_8"O"_7) xx overbrace(cancel("1 mol C"_6"H"_8"O"_7)/(192 cancel("g C"_6"H"_8"O"_7)))^"Molar mass of citric acid" xx ("3 mol CO"_2)/cancel("1 mol C"_6"H"_8"O"_7)#
#= "0.181 mols CO"_2(g)#
That is how many mols of
#0.181 cancel("mols CO"_2) xx overbrace("44 g CO"_2/cancel("1 mol CO"_2))^"Molar mass of carbon dioxide"#
#= ul("7.975 g CO"_2(g))#
From the
