# For the following reaction, what is the maximum mass of "HCl" that can be obtained by the reaction "37.5 g" "BCl"_3 and "60.0 g" of "H"_2"O"?

## $\text{BCl"_3 (g) + 3"H"_2"O"(l) -> "H"_3"BO"_3(s) + 3"HCl} \left(g\right)$

Jun 13, 2017

$\text{35.0 g HCl}$

#### Explanation:

Start by determining which reactant, if any, acts as a limiting reagent.

You know from the balanced chemical equation

${\text{BCl"_ (3(g)) + color(blue)(3)"H"_ 2"O"_ ((l)) -> "H"_ 3"BO"_ (3(s)) + 3"HCl}}_{\left(g\right)}$

that every mole of boron trichloride consumes $\textcolor{b l u e}{3}$ moles of water, so use the molar masses of the two reactants to convert the masses to moles.

37.5 color(red)(cancel(color(black)("g"))) * "1 mole BCl"_3/(117.17color(red)(cancel(color(black)("g")))) = "0.32005 moles BCl"_3

60.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "3.3306 moles H"_2"O"

So, pick one of the two reactants and check to see if you have enough moles of the other reactant to ensure that all the moles of the first reactant take part in the reaction.

For boron trichloride, you have

0.32005color(red)(cancel(color(black)("moles BCl"_3))) * (color(blue)(3)color(white)(.)"moles H"_2"O")/(1color(red)(cancel(color(black)("mole BCl"_3)))) = "0.96015 moles H"_2"O"

So, in order for all the moles of boron trichloride to take part in the reaction, you need $0.96015$ moles of water. Since you have more than enough moles of water to ensure that this happens

overbrace("3.3306 moles H"_2"O")^(color(blue)("what you have")) " " > " " overbrace("0.96015 moles H"_2"O")^(color(blue)("what you need"))

you can say that water is in excess, which implies that boron trichloride acts as a limiting reagent, i.e. it gets completely consumed before all the moles of water get the chance to react.

You now know that $0.32005$ moles of boron trichloride take part in the reaction, which implies that the reaction produces

0.32005 color(red)(cancel(color(black)("moles BCl"_3))) * "3 moles HCl"/(1color(red)(cancel(color(black)("mole BCl"_3)))) = "0.96014 moles HCl"

To convert this to grams, use the molar mass of hydrogen chloride

$0.96015 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles HCl"))) * "36.461 g"/(1color(red)(cancel(color(black)("mole HCl")))) = color(darkgreen)(ul(color(black)("35.0 g}}}}$

The answer is rounded to three sig figs.