A 20.00 mL of 0.200mol/L magnesium chloride solution is added to a 95.00 mL of 0.800 silver (I) nitrate solution, and a precipitate is formed.

a.Determine the concetration of the nitrate ions at the end of the reaction
I was able to calculate the things and make an IRF table but i don't know what to do after that plz help!

millimoles MgCl2 = 20 x 0.200 = 4.0

millimoles AgNO3 = 95 x 0.8 = 7.6

.........2AgNO3 + MgCl2 ==> 2AgCl + Mg(NO3)2
The nitrate ions never enter any reaction so they are at the end what they were initially. They were 0.8M but that's been diluted by 95/(95+20)
That is 0.8M x (95/(95+20) = ?

To determine the concentration of nitrate ions at the end of the reaction, you need to use the stoichiometry of the balanced chemical equation and the information provided about the volumes and concentrations of the reactant solutions.

First, let's write the balanced chemical equation for the reaction between magnesium chloride and silver nitrate:

MgCl2 + 2AgNO3 → 2AgCl + Mg(NO3)2

According to the equation, one mole of magnesium chloride reacts with two moles of silver nitrate to produce two moles of silver chloride and one mole of magnesium nitrate.

Next, let's calculate the number of moles of silver nitrate used in the reaction. We can do this using the volume and concentration information:

Volume of silver nitrate solution = 95.00 mL = 0.09500 L
Concentration of silver nitrate solution = 0.800 mol/L

Moles of silver nitrate = Volume × Concentration
= 0.09500 L × 0.800 mol/L
= 0.0760 mol

Since two moles of silver nitrate react with one mole of magnesium nitrate, the number of moles of magnesium nitrate formed is also 0.0760 mol.

Now, we need to consider the total volume of the resulting solution. The volume will be the sum of the volumes of the magnesium chloride and silver nitrate solutions:

Volume of magnesium chloride solution = 20.00 mL = 0.02000 L
Total volume = 0.02000 L + 0.09500 L = 0.11500 L

To find the concentration of nitrate ions, we divide the moles of magnesium nitrate by the total volume:

Concentration of nitrate ions = Moles of magnesium nitrate / Total volume
= 0.0760 mol / 0.11500 L
≈ 0.661 mol/L

Therefore, the concentration of nitrate ions at the end of the reaction is approximately 0.661 mol/L.