When 500. mL of a 1.8 M solution of X is combined with 500. mL of a 1.8 M solution of Y, the resulting solution has a concentration of 0.60 M Y and 0.60 M Z. No more of substance X remains in the flask.

1. How many moles each of X and Y are present before the reaction occurs?

2. How many moles each of Y and Z are present after the reaction occurs?

3. How many moles each of X and Y have reacted?

4. What is the balanced equation for this reaction?

Consider the reaction between two solutions, X and Y, to produce substance Z: aX + bY → cZ 

Answer:

(1) 0.90 mol of X and Y are present before the reaction occurs.

(2) 0.60 mol of Y and Z are present after the reaction occurs.

(3) 0.90 mol of X and 0.30 mol of Y have reacted.

(4) ?3X+Y→2Z?

Step-by-step explanation

Molarity of a solution = (Number of moles of solute)/(Volume of solution in Liter)

(1) Number of mol of X in 500. mL (0.500 L) of a 1.8 M solution of X = ?(1.8×0.500)? mol = 0.90 mol

Number of mol of Y in 500. mL (0.500 L) of a 1.8 M solution of Y = ?(1.8×0.500)? mol = 0.90 mol

So, 0.90 mol of X and Y are present before the reaction occurs.

(2) Total volume of reaction mixture = (500.+500.) mL = 1000 mL = 1.00 L

Number of mol of Y in 1.00 L of 0.60 M solution of Y = ?(0.60×1.00)? mol = 0.60 mol

Number of mol of Z in 1.00 L of 0.60 M solution of Z = ?(0.60×1.00)? mol = 0.60 mol

So, 0.60 mol of Y and Z are present after the reaction occurs.

(3) As X is completely consumed during the reaction therefore 0.90 mol of X has reacted.

Number of mol of Y has reacted = (Initial number of mol of Y)-(Final number of mol of Y) = (0.90 mol)-(0.60 mol) = 0.30 mol

(4) 0.30 mol of Y reacts with 0.90 mol of X and produces 0.60 mol of Z

Number of mol of X : Number of mol of Y : Number of mol of Z

= 0.90 : 0.30 : 0.60

= 3:1:2

Balanced reaction: ?3X+Y→2Z?