Question

What is the maximum amount of HCl, in grams, that can be produced if 60.0 g of BCl3 and 37.5 g of H2O are reacted according to the following balanced reaction?
BCl3 + 3 H2O → H3BO3 + 3 HCl

Answer 1

- Write and balance the equation
- Find the limiting and excess reagents.
- Consider the molar/volume/mass ratios.

For this exercise, we can deal with mass ratios.
Mr of the following substances are:
BCl3 - 117.3g
H2O - 18g
HCL- 36.5g

Equating with moles, you can see that according to the equation, for every 117.3g of BCl3, we have (3*18) or 54g of H2O.
Now, according to this proportionality , let us use the amounts they gave us to see which one has is the limiting and excess reagents, respectively.

If we have 60g of BCl3, we were suppose to have \[\frac{ 60*54 }{ 117.3 }=27.6g\] of H2O, but we have more than that, 37.5g.
Meaning, H2O is in excess and BCl3 is the limiting reagent.
For this exercises, we work out using the data from the limiting reagents.

Therefore, we now deal with BCl3 and HCl. according to the equation, for every 117.3g of BCl3, we have (3*36.5) or 109.5g of HCl.
Then if we are given 60g of BCl , we are expected to have:\[\frac{ 60*109.5 }{ 117.3 }=56.01g\] of HCl.

Answer 2

Thank you!

Answer 3

No problem.