Question

Diborane, B2H6 a possible rocket propellant, can be made by using lithium hydride (LiH): 6 LiH+ 2 BCl2àB2H6+ 6 LiCl . If you mix 200 lb of LiH with 1000 lb of BCl3 , you recover 45 lb of B2H6. Determine (a) Limiting reactant (b) The excess reactant (c) The percent excess reactant (d) The percent conversion of LiH to B2H6 (e) Lb of LiCL produced

Answer 1

(a) Limiting reactant =
`LiH`

(b) The excess reactant =
`BCl_3`

(c) The percent of excess reactant is, 50.87 %

(d) The percent yield of
`B_2H_6` or percent conversion of
`LiH` to
`B_2H_6` is, 38.80 %

(e) The mass of
`LiCl` produced is, 1066.42 lb

Explanation : Given,

Mass of
`LiH` = 200 lb = 90718.5 g

conversion used : (1 lb = 453.592 g)

Mass of
`BCl_3` = 1000 lb = 453592 g

Molar mass of
`LiH` = 7.95 g/mole

Molar mass of
`BCl_3` = 117.17 g/mole

Molar mass of
`B_2H_6` = 27.66 g/mole

Molar mass of
`LiCl` = 42.39 g/mole

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

`\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=(90718.5g)/(7.95g/mole)=11411.13moles`

`\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=(453592g)/(117.17g/mole)=3871.23moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`6LiH+2BCl_3\rightarrow B_2H_6+6LiCl`

From the balanced reaction we conclude that

As, 6 moles of
`LiH` react with 1 mole of
`BCl_3`

So, 11411.13 moles of
`LiH` react with
`(11411.13)/(6)=1901.855` moles of
`BCl_3`

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

Moles of remaining excess reactant = 3871.23 - 1901.855 = 1969.375 moles

Total excess reactant = 3871.23 moles

Now we have to determine the percent of excess reactant
`(BCl_3)`.

`\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}* 100`

`\% \text{ excess reactant}=(1969.375)/(3871.23)* 100=50.87\%`

The percent of excess reactant is, 50.87 %

Now we have to calculate the moles of
`B_2H_6`.

As, 6 moles of
`LiH` react to give 1 mole of
`B_2H_6`

So, 11411.13 moles of
`LiH` react to give
`(11411.13)/(6)=1901.855` moles of
`B_2H_6`

Now we have to calculate the mass of
`B_2H_6`.

`\text{Mass of }B_2H_6=\text{Moles of }B_2H_6* \text{Molar mass of }B_2H_6`

`\text{Mass of }B_2H_6=(1901.855mole)* (27.66g/mole)=52605.3093g`

Now we have to calculate the percent yield of
`B_2H_6`.

`\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}* 100=(20411.7g)/(52605.3093g)* 100=38.80\%`

The percent yield of
`B_2H_6` or percent conversion of
`LiH` to
`B_2H_6` is, 38.80 %

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

As, 6 moles of
`LiH` react to give 6 mole of
`LiCl`

So, 11411.13 moles of
`LiH` react to give 11411.13 moles of
`LiCl`

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

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

`\text{Mass of }LiCl=(11411.13mole)* (42.39g/mole)=483717.8007g=1066.42lb`

The mass of
`LiCl` produced is, 1066.42 lb