Chrome just got better! A new version is available. Update a. How many grams of

Question

Chrome just got better! A new version is available. Update a. How many grams of the first product can he produced from 20.0 grams of first reactant and 20.0 grams of the second reactant? b. Which reactant is the limiting reactant? c. How many grams of the excess reactant are reacted? d. How many grams of the excess reactant are left unreacted? 11. A student carried out the reaction shown by reacting 10.0g P4 with excess CI2, and obtained 35.0 g of PCI3. P4 + 6Cl2, right arrow 4PCl3 a. What is the theoretical yield of PCI3? b. Calculate the % yield of PCl3.

Explanation / Answer

11.

The chemical reaction is P4 + 6Cl2 ----------> 4PCl3

Given, 10.0g of P4

Moles of P4 = 10.0 / 123.8 = 0.081 mol.

From the chemical equation, it is clear that 1 mol of P4 gives 4 mol PCl3.

so, 0.081 mol of P4 gives 4 x 0.081 = 0.324 mol PCl3

So, mass of PCl3 = Moles x Mass

= 0.324 x 137.33

= 44.49 g.

So, theoretical yield = 44.49 g

Practical yield = 35.0 g

Pecentage yield = (Practical yield / theoretical yield) x 100

= (35.0 / 44.49 ) x 100

= 78.66%

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