A 3.90 g sample of Cl2 reacts with PCl3 to form 11.45 g of PCl5 according to the reaction below. How much PCl3 is needed? [Hint: You do not need to calculate moles to answer.]

PCl3 + Cl2 —> PCl5 Amount of PCl3:

Show work here

We need 7.55 grams of PCl3

Explanation:

Step 1: Data given

Mass of Cl2 = 3.90 grams

Mass of PCl5 = 11.45 grams

Molar mass Cl2 = 70.9 g/mol

Molar mass PCl5 = 208.24 g/mol

Step 2: The balanced equation

PCl3 + Cl2 → PCl5

Step 3: calculate moles

Moles = mass / molar mass

Moles Cl2 = 3.90 grams / 70.9 g/mol

Moles Cl2 = 0.0550 moles

Moles PCl5 = 11.45 grams /208.24 g/mol

Moles PCl5 = 0.0550 moles

Step 4: Calculate moles PCl3

For 1 mol PCl3 we need 1 mol Cl2 to produce 1 mol PCl5

For 0.0550 moles PCl5 we will need 0.0550 moles Cl2 and 0.0550 moles PCl3

Step 5: Calculate mass PCl3

Mass PCl3 = moles * molar mass

Mass PCl3 = 0.0550 moles * 137.33 g/mol

Mass PCl3 = 7.55 grams

Without calculating the number of moles:

Mass of Cl2 + mass PCl3 = mass PCl5

Mass PCl3 = 11.45 - 3.90 = 7.55 grams

We need 7.55 grams of PCl3

yes it is balanced