Respuesta :
Answer :
(a) Limiting reactant = [tex]LiH[/tex]
(b) The excess reactant = [tex]BCl_3[/tex]
(c) The percent of excess reactant is, 50.87 %
(d) The percent yield of [tex]B_2H_6[/tex] or percent conversion of [tex]LiH[/tex] to [tex]B_2H_6[/tex] is, 38.80 %
(e) The mass of [tex]LiCl[/tex] produced is, 1066.42 lb
Explanation : Given,
Mass of [tex]LiH[/tex] = 200 lb = 90718.5 g
conversion used : (1 lb = 453.592 g)
Mass of [tex]BCl_3[/tex] = 1000 lb = 453592 g
Molar mass of [tex]LiH[/tex] = 7.95 g/mole
Molar mass of [tex]BCl_3[/tex] = 117.17 g/mole
Molar mass of [tex]B_2H_6[/tex] = 27.66 g/mole
Molar mass of [tex]LiCl[/tex] = 42.39 g/mole
First we have to calculate the moles of [tex]LiH[/tex] and [tex]BCl_3[/tex].
[tex]\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=\frac{90718.5g}{7.95g/mole}=11411.13moles[/tex]
[tex]\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{453592g}{117.17g/mole}=3871.23moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]6LiH+2BCl_3\rightarrow B_2H_6+6LiCl[/tex]
From the balanced reaction we conclude that
As, 6 moles of [tex]LiH[/tex] react with 1 mole of [tex]BCl_3[/tex]
So, 11411.13 moles of [tex]LiH[/tex] react with [tex]\frac{11411.13}{6}=1901.855[/tex] moles of [tex]BCl_3[/tex]
From this we conclude that, [tex]BCl_3[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]LiH[/tex] 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 [tex](BCl_3)[/tex].
[tex]\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100[/tex]
[tex]\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%[/tex]
The percent of excess reactant is, 50.87 %
Now we have to calculate the moles of [tex]B_2H_6[/tex].
As, 6 moles of [tex]LiH[/tex] react to give 1 mole of [tex]B_2H_6[/tex]
So, 11411.13 moles of [tex]LiH[/tex] react to give [tex]\frac{11411.13}{6}=1901.855[/tex] moles of [tex]B_2H_6[/tex]
Now we have to calculate the mass of [tex]B_2H_6[/tex].
[tex]\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6[/tex]
[tex]\text{Mass of }B_2H_6=(1901.855mole)\times (27.66g/mole)=52605.3093g[/tex]
Now we have to calculate the percent yield of [tex]B_2H_6[/tex].
[tex]\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%[/tex]
The percent yield of [tex]B_2H_6[/tex] or percent conversion of [tex]LiH[/tex] to [tex]B_2H_6[/tex] is, 38.80 %
Now we have to calculate the moles of [tex]LiCl[/tex].
As, 6 moles of [tex]LiH[/tex] react to give 6 mole of [tex]LiCl[/tex]
So, 11411.13 moles of [tex]LiH[/tex] react to give 11411.13 moles of [tex]LiCl[/tex]
Now we have to calculate the mass of [tex]LiCl[/tex].
[tex]\text{Mass of }LiCl=\text{Moles of }LiCl\times \text{Molar mass of }LiCl[/tex]
[tex]\text{Mass of }LiCl=(11411.13mole)\times (42.39g/mole)=483717.8007g=1066.42lb[/tex]
The mass of [tex]LiCl[/tex] produced is, 1066.42 lb
