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

ANSWERS

2016-09-28 06:12:16

Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is. First, how many moles of each substance are there the molar mass of BCl3 is 117.17 grams so 37.5 g / 117.17 is ~ .32 mol. The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol. Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles. For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles. But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl. .96 x 36.46 = ~35 g

ADD ANSWER