Question

When phosphorus reacts with chlorine, phosphorus trichloride is formed according to the following equation: ___P4(s) ___Cl2(g) → ___PCl3(l) (unbalanced) Balance the equation and determine how many grams of chlorine would be required to react with 21.2 g of phosphorus.

Answer 1

Answer:

P4 + 6Cl2 ------> 4PCl3

145.59 g of Cl.

Explanation:

First to all, let's balance the equation. The principle for balance an equation, try to see this like a balance of two plates. You need to have the same numbers of atoms of both sides of the plates to equilibrate the balance.

In this case, we have 4 atoms of P, 2 of Cl on one side, and for the other side we have 1 atom of P and 3 of chlorine.

To balance this, we balance first the P on the right side putting a leading 4, and we have 4PCl3.

Now this 4, is also added to the Cl, and it's multiplied, so we have 12 Cl in total on the right side. To balance this, we put a leading 6 on the left side in the Cl.

The final balanced equation would be

P4 + 6Cl2 ------> 4PCl3

Now that we have the balanced equation, let's calculate the grams of Cl needed:

In this case, first, let's calculate the mole of P used in the 21.2 g. We need the atomic weight of P, which is 30.97 g/mol and the Cl is 35.45 g/mol

moles of P = g/AW = 21.2 / 30.97 = 0.6845 moles

With the balanced equation, we can do a relation between P and Cl.

If 1 mole P reacts with 6 moles of Cl,

then 0.6845 mole P reacts with X moles of Cl.

Solving for X:

X = moles of Cl = 0.6845 * 6 = 4.107 moles

Finally for the grams of Cl:

g = 4.107 * 35.45 = 145.59 g of Cl.