Chemistry, 23.08.2019 05:30 epelmanr

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

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Answer from: makayladurham19

Answer :

(a) Limiting reactant = LiH

(b) The excess reactant = BCl_3

(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 .

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

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

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 $\frac{11411.13}{6}=1901.855$ 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) .

$\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100$

$\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%$

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 $\frac{11411.13}{6}=1901.855$ moles of B_2H_6

Now we have to calculate the mass of B_2H_6 .

$\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6$

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

Now we have to calculate the percent yield of B_2H_6 .

$\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%$

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 .

$\text{Mass of }LiCl=\text{Moles of }LiCl\times \text{Molar mass of }LiCl$

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

The mass of LiCl produced is, 1066.42 lb

Answer from: Quest

answer: e. sour taste

explanation: acids are those substances which dissociates in water to give $h^+$ ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

$hcl\rightarrow h^++cl^-$

bases are those substances which dissociates in water to give $oh^-$ ions or accepts $h^+$ ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.

$naoh\rightarrow na^++oh^-$

$nh_3+h^+\rightarrow nh_4^+$

Answer from: Quest

with this one i know the answer in craters one reason i looked it up and i learned about it there is my evidence

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