Question

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced reaction. A possibly useful molar mass is BC13 117.16 g/mol. BC13(g)+3 H20(1) -- H3BO3(s)+3 HC1(g)

Answer 1

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of
`BCl_3` = 60 g

Mass of
`H_2O` = 37.5 g

Molar mass of
`BCl_3` = 117 g/mole

Molar mass of
`H_2O` = 18 g/mole

Molar mass of
`HCl` = 36.5 g/mole

First we have to calculate the moles of
`BCl_3` and
`H_2O`.

`\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=(60g)/(117g/mole)=0.513moles`

`\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=(37.5g)/(18g/mole)=2.083moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)`

From the balanced reaction we conclude that

As, 1 mole of
`BCl_3` react with 3 mole of
`H_2O`

So, 0.513 moles of
`BCl_3` react with
`3* 0.513=1.539` moles of
`H_2O`

From this we conclude that,
`H_2O` is an excess reagent because the given moles are greater than the required moles and
`BCl_3` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`HCl`.

As, 1 mole of
`BCl_3` react to give 3 moles of
`HCl`

So, 0.513 moles of
`BCl_3` react to give
`3* 0.513=1.539` moles of
`HCl`

Now we have to calculate the mass of
`HCl`.

`\text{Mass of }HCl=\text{Moles of }HCl* \text{Molar mass of }HCl`

`\text{Mass of }HCl=(1.539mole)* (36.5g/mole)=56.1735g`

Therefore, the theoretical yield of HCl is, 56.1735 grams