Question

An electric furnace produces phosphorus by the following reaction:

${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\left(\mathrm{s}\right)+5\mathrm{C}\left(\mathrm{s}\right)+3{\mathrm{SiO}}_2\left(\mathrm{s}\right)\to 3{\mathrm{CaSiO}}_3\left(\mathrm{s}\right)+5\mathrm{CO}\left(\mathrm{g}\right)+2\mathrm{P}\left(\mathrm{l}\right)$

An initial reaction mixture contains 1500 kg calcium phosphate, 250 kg carbon, and $1.0\times {10}^{3}kg$${\mathrm{SiO}}_2$.

1. What is the limiting reagent?
2. What is the theoretical yield of phosphorus?
3. After reaction the slag (solid residue) was analyzed. It contained role="math" localid="1650271274566" $3.8\%\text{C}$, $5.8\%\text{P}$, and $26.6\%\mathrm{Ca}$ by mass. What was the actual yield of phosphorus in kg? What was the percent yield?

Short answer

1. carbon is the limiting reagent
2. The theoretical yield of phosphorus is $258\text{kg}$.
3. The actual yield of phosphorus is $172\text{kg}$and thepercent yield is$67\%$.

Step-by-step solution

Definition

A limiting reagent in a reaction is the reactant or substance that is totally consumed during the reaction's conclusion. The amount of the limiting reagent employed in the reaction determines the total amount of products generated. When a certain amount of reacting substances is supplied, the amount of product created by the limiting reagent is the smallestsince it is consumed fully first in the reaction.

Determination of amount of phosphorous by 1500kg Ca3PO42.

From the given, the balanced chemical equation is follows,

${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\left(\mathrm{s}\right)+5\mathrm{C}\left(\mathrm{s}\right)+3{\mathrm{SiO}}_2\left(\mathrm{s}\right)\to 3{\mathrm{CaSiO}}_3\left(\mathrm{s}\right)+5\mathrm{CO}\left(\mathrm{g}\right)+2\mathrm{P}\left(\mathrm{l}\right)$

The amount of phosphorous generated by a given number of reactants may be estimated individually as follows:

a)

1) In the case of 1500 kg ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$:

When one mole of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$ is is reacted, the amount of phosphorous created is 2 mole, according to the above balanced equation. The molar weights of phosphorous and calcium phosphate are $30.97\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$and $310.10\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$respectively, $310.18\mathrm{g}$calcium phosphate produces $61.95\mathrm{g}$or $\left(30.97\mathrm{g}x2\right)$ phosphorous. So, when 1500 kg calcium phosphate is reacted, the quantity of phosphorous created may be computedas follows:

Mass of P producedrole="math" localid="1650272629835" $=\left(\frac{\left(2\times 30.97\right)\mathrm{g}\mathrm{P}}{310.10\mathrm{g}{\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2}\right)\left(1500\mathrm{kg}{\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\right)\left(\frac{1000\mathrm{g}}{1\mathrm{kg}}\right)$

$=299613\mathrm{g}\mathrm{P}$

$=299613\mathrm{g}\mathrm{P}\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)$

$=300\mathrm{kg}\mathrm{P}$

Determination of amount of phosphorous by 250kg C.

2)

When 5 moles of carbon are reacted, the quantity of phosphorous created is 2 moles, according to the above balanced equation. The molar weights of phosphorous and carbon are $30.97\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$and $12.01\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$respectively, $60.05\mathrm{g}$or role="math" localid="1650273408990" $\left(12.01\mathrm{g}x\text{5}\right)$carbon produces $61.94\mathrm{g}$or $\left(30.97\mathrm{g}x\text{2}\right)$phosphorous. As a result, when 250 kg carbon is consumed, the quantity of phosphorous created may be estimated as follows:

Mass of P produced$=\left(\frac{\left(2\times 30.97\right)\mathrm{g}\mathrm{P}}{\left(5\times 12.01\right)\mathrm{g}\mathrm{C}}\right)\left(250\mathrm{kg}\mathrm{C}\right)\left(\frac{1000\mathrm{g}}{1\mathrm{kg}}\right)$

$=257868.44\mathrm{g}\mathrm{P}$

$\text{=257868.44gP}\left(\frac{1kg}{1000g}\right)$

$=258\mathrm{kg}\mathrm{P}$

Determination of amount of phosphorous by 1.0×103kg SiO2.

3)

When 3 moles of silicon dioxide are reacted, the quantity of phosphorous created is 2 moles, according to the above balanced equation. The molar weights of phosphorous and carbon are $30.97\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$and $60.09\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$respectively, $180.27\mathrm{g}$or $\left(60.09\mathrm{g}x\text{3}\right)$carbon produces $61.94\mathrm{g}$or $\left(30.97\mathrm{g}\mathrm{x}\text{2}\right)$phosphorous. As a result, when $1.0\times {10}^3\mathrm{kg}$ silicon dioxide is consumed, the quantity of phosphorous created may be estimated as follows:

Mass of P produced$=\left(\frac{\left(2\times 30.97\right)\mathrm{g}\mathrm{P}}{\left(3\times 60.09\right)\mathrm{g}{\mathrm{SiO}}_2}\right)\left(1.0\times {10}^3\mathrm{kg}{\mathrm{SiO}}_2\right)\left(\frac{1000\mathrm{g}}{1\mathrm{kg}}\right)$

$=343595.72\mathrm{g}\mathrm{P}$

$=343595.72\mathrm{g}\mathrm{P}\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)$

$=343.60\mathrm{kg}\mathrm{P}$

According to the calculations above, the quantity of phosphorous created in the case of carbon is the least. As a result, carbon is the limiting reagent.

Determination of theoretical yield of phosphorous

b)

Because carbon is discovered to be the limiting reagentin the given reaction, the theoretical yield of the reaction is the yield of phosphorous generated by the given quantity of carbon. So, based on the calculations above, the theoretical phosphorous production is 258 kg.

Determination of total mass of the solid reactants

c)

To begin, calculate the total mass of the solid reactantsas follows:

Total mass of initial reactants$=\mathrm{mass}\mathrm{of}{\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2+\mathrm{mass}\mathrm{of}\mathrm{C}+\mathrm{mass}\mathrm{of}{\mathrm{SiO}}_2$

$=1500\mathrm{kg}+250\mathrm{kg}+1.0\times {10}^3\mathrm{kg}$

$=2750\mathrm{kg}$

The products are carbon monoxide, a gas, and liquid phosphorous, both of which will reduce the quantity of solid leftovers generated. Let x be the kg mol of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$that interacted, and the balanced equation be:

${\mathrm{xCa}}_3{\left({\mathrm{PO}}_4\right)}_2\left(\mathrm{s}\right)+5\mathrm{xC}\left(\mathrm{s}\right)+3{\mathrm{xSiO}}_2\left(\mathrm{s}\right)\to 3{\mathrm{xCaSiO}}_3\left(\mathrm{s}\right)+5\mathrm{xCO}\left(\mathrm{g}\right)+2\mathrm{xP}\left(\mathrm{l}\right)$

Determination of the amount of C in kg moles.

The amount of C utilised will determine the yield because it is the limiting reagent. The kilogramme moles of carbon in 250 kilogram of carbon may be computed as follows:

$\mathrm{kg}\mathrm{mol}\mathrm{C}=\left(250\mathrm{kg}\mathrm{C}\right)\left(\frac{1\mathrm{kg}\mathrm{mol}\mathrm{C}}{12.01\mathrm{kg}\mathrm{C}}\right)$

$\text{= 20.82}\mathrm{kg}\mathrm{mol}\mathrm{C}$

Because the slag contains 3.8% carbon,the quantity of carbon reacted in kg moles may be calculated as follows:

$\mathrm{mass}\%\mathrm{of}\mathrm{C}=\left(\frac{\text{mass}\mathrm{of}\mathrm{C}\mathrm{reacted}}{\mathrm{mass}\mathrm{of}\mathrm{Slag}}\right)100$

$3.8=\left(\frac{\text{2}\left(0.8\mathrm{kmol}\mathrm{C}-5\mathrm{x}\right)\text{12.01}\mathrm{kg}{\displaystyle \raisebox{1ex}{$\mathrm{C}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{kmol}$}\right.}\mathrm{C}}{2750\mathrm{kg}-\mathrm{mass}\mathrm{of}\mathrm{CO}-\mathrm{mass}\mathrm{of}\mathrm{P}}\right)100$

$3.8=\left(\frac{\text{2}\left(0.8\mathrm{kmol}\mathrm{C}-5\mathrm{x}\right)\text{12.01}\mathrm{kg}{\displaystyle \raisebox{1ex}{$\mathrm{C}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{kmol}$}\right.}\mathrm{C}}{2750\mathrm{kg}-5\mathrm{x}\left(\text{28.01}\mathrm{kg}{\displaystyle \raisebox{1ex}{$\mathrm{C}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{kmol}$}\right.}\mathrm{CO}\right)-2\mathrm{x}\left(\text{30.97}\mathrm{kg}{\displaystyle \raisebox{1ex}{$\mathrm{P}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{kmlo}$}\right.}\mathrm{P}\right)}\right)100$

$\left(20.8-5\mathrm{x}\right)\text{12.01}=0.038\left[2750-5\mathrm{x}\left(28.01\right)-2\mathrm{x}\left(30.97\right)\right]$

$249.808-6005\mathrm{x}=104.5-5.322\mathrm{x}-2.354\mathrm{x}$

$\mathrm{x}=2.77$

So, by inserting the value of x in 5x, the kilo moles of C reacted may be calculated:

Moles of reacted$=\left(5\mathrm{x}\right)\mathrm{kmol}$

$=5\left(2.77\right)\mathrm{kmol}$

$=13.85\mathrm{kmol}$

Determination of the actual yield of phosphorous in kg

As a result, 13.85 kilo mol of carbonwas really reacted. Because 5 moles of carbon generate 2 moles of phosphorous, the quantity of phosphorous produced by 13.85 k mol carbon is calculated as follows:

Actual yield of $\mathrm{P}=\left(13.85\mathrm{kmol}\mathrm{C}\right)\left(\frac{2\mathrm{kmol}\text{P}}{5\mathrm{kmol}\mathrm{C}}\right)\left(\frac{30.97\mathrm{kg}\text{P}}{1\mathrm{kmol}\mathrm{P}}\right)$

$=172\mathrm{kg}$

Thus the actual yield of phosphorous is $172\mathrm{kg}\mathrm{P}$.

Determination of the percentage yield of phosphorous.

The theoretical P yield was 258 kg; however the actual yield was just 172 kg. As a result, P's percentage yield is computed as follows:

Percentage yield of $\mathrm{P}=\left(\frac{\mathrm{Actual}\mathrm{yield}}{\mathrm{theoretical}\mathrm{yield}}\right)100$

$=\left(\frac{172\mathrm{kg}}{258\mathrm{kg}}\right)100$

$=67\%$

Therefore, the percentage yield of phosphorous is$67\%$.