Question

What is the percent yield for the reaction pcl3(g) + cl2(g) ? pcl5(g) if 119.3 g of pcl5 (? = 208.2 g/mol) are formed when 61.3 g of cl2 (? = 70.91 g/mol) react with excess pcl3? a 66.3%?

Answer 1

The percent yield for the reaction is 66.3%

• The actual yield refers to the amount of product that is gotten from a chemical reaction
• Theoretical yield refers to the amount of product that is gotten from the balanced equation. The derived value is obtained using limiting reactant to find the product.
• The units for the actual yield and theoretical yield are the same. It can be moles of grams

Further Explanation

The equation for percent yield is as follows:

Percent yield = (actual yield/theoretical yield) x 100%

From the given question, the complete equation is as follows:

pcl₃ (g) + cl₂ (g) = pcl₅ (g)

Firstly, you have to calculate the theoretical yield  

Theoretical yield = 61.3 g Cl₂ (1 mol / 70.91 g) (1 mol PCl₅ / 1 mol Cl₂) (208.2 g / 1 mol)  

For proper clarification

= 61.3 / 70.91 x 208.2

= 180.01 g Pcl₅

Therefore, to calculate the percent yield, you have to divide the valued derived (theoretical yield) with the actual yield and multiply it by 100

Therefore, you have:

Percent yield = (actual yield/theoretical yield) x 100%

Substituting the value:

Percent yield = 119.3 g / 180.01g x 100

= 66.283

= 66.3%

LEARN MORE:

• How many kilojoules are absorbed when 280 g of pcl5 reacts? pcl5(g)→pcl3(g)+cl2(g)δh=+67kj?  brainly.com/question/6420309
• What mass of PCl5 is expected from the reaction of 73.7 g PCl3 with excess chlorine? brainly.com/question/12741054

KEYWORDS:

• percent yield
• theoretical yield
• chemical reaction
• moles
• grams

Answer 2

The balanced chemical reaction would be 

pcl3(g) + cl2(g) = pcl5(g)

First, we need to determine the theoretical yield of the reaction assuming that all of the limiting reactant is used up. Then, we divide the actual yield to the theoretical yield to determine the percent yield.

Theoretical yield = 61.3 g Cl2 ( 1 mol / 70.9 g) ( 1 mol PCl5 / 1 mol Cl2 ) ( 208.2 g / 1 mol ) = 180.01 g PCl5

Percent yield = 119.3 g / 180.01 g x100 = 66.3%