Question

P4O10 + 6PCl5---> 10 POCl3

How many grams of POCl3 are produced when 225.0g of P4O10 and 675.0g of PCl5 react?

Answer 1

When 225.0g of P4O10 and 675.0g of PCl5 react, the limiting reactant is PCl5. The mass of POCl3 produced is 829.05g.

Step-by-step explanation:

To find the amount of POCl3 produced when 225.0g of P4O10 and 675.0g of PCl5 react, we need to determine the limiting reactant first.

To do so, we convert the given masses of P4O10 and PCl5 to moles using their molar masses. The balanced chemical equation tells us that 1 mole of P4O10 produces 10 moles of POCl3, while 1 mole of PCl5 produces 10 moles of POCl3.

Comparing the moles of product produced from each reactant will allow us to determine the limiting reactant. After determining the limiting reactant, we can use stoichiometry to find the mass of POCl3 produced.

Step 1: Convert grams to moles for P4O10 and PCl5
1 mole P4O10 = 283.88 g P4O10
1 mole PCl5 = 208.23 g PCl5
Moles of P4O10 = 225.0 g P4O10 × (1 mol P4O10 / 283.88 g P4O10) = 0.7929 mol P4O10
Moles of PCl5 = 675.0 g PCl5 × (1 mol PCl5 / 208.23 g PCl5) = 3.2436 mol PCl5

Step 2: Determine the limiting reactant
Using the balanced chemical equation: P4O10 + 6PCl5 → 10POCl3
1 mole P4O10 produces 10 moles POCl3
1 mole PCl5 produces 10 moles POCl3
Moles of POCl3 from P4O10 = 0.7929 mol P4O10 × (10 mol POCl3 / 1 mol P4O10) = 7.929 mol POCl3
Moles of POCl3 from PCl5 = 3.2436 mol PCl5 × (10 mol POCl3 / 6 mol PCl5) = 5.406 mol POCl3

Since the moles of POCl3 from PCl5 (5.406 mol) are less than the moles of POCl3 from P4O10 (7.929 mol), PCl5 is the limiting reactant.

Step 3: Calculate the mass of POCl3 produced
Molar mass of POCl3 = 153.33 g/mol
Mass of POCl3 = 5.406 mol POCl3 × (153.33 g POCl3 / 1 mol POCl3) = 829.05 g POCl3

Answer 2

P4O10 + 6PCl5 --> 10 POCl3

That equation is balances (same number of atoms of each element on the left than on the right).

1) Writhe the stechiometric proportions:

1 mol P4O10 : 6mol PCl5 : 10 mol POCl3

2) Calculate the molar masses of each compound (using the aomic masses of the elements)

molar mass of P4O10 = 4*31g/mol + 10*16g/mol = 284 g/mol

molar mass of PCl5 = 31g/mol * 5*35.5 g/mol = 173 g/mol

molar mass of POCl3 = 31 g/mol + 16g/mol + 3*35.5g/mol = 153.5 g/mol

3) Pass the information in grams to mol by dividing grams / molar mass

225.0 g P4O10 / 284 g/mol = 0.792 mol P4O10

675.0 g PCl5 / 173 g/mol = 3.902 mol PCl5

4) Find the limitant reagent

ratio 3.902 mol PCl5 / 0.792 mol P4O10 = 4.93

4.93 < 6/1 which is the theoretical ratio, then there is less PCl5 and it will be consumed before ending the P4O10.

5) Use the # of moles of the limitan reagent to find the # of moles of the product, using the theoretical proportions

[10 mol POCl3 / 6mol PCl5 ] * 3.902 mol PCl5 =6.50 mol POCl3

6) Pass the number of moles of POCl3 to grams, using its molar mass

6.50 mol POCl3 * 153.5 g/mol = 997.8 grams

Answer: 997.8 g