if 22 mL of 0.143M MgCl2 is needed to completely react 15mL of AgNO3 solution, what is the molarity of AgNO3

042 M

Explanation:

1) Write the molecular chemical equation

MgCl₂ (aq) + 2AgNO₃ (aq) → 2AgCl (s) + Mg (NO₃) ₂ (aq)

2) Mole ratios

1 mol MgCl₂ : 2 mol AgNO₃ : 2 mol AgCl : 1 mol Mg (NO₃) ₂

3) Number of moles of MgCl₂

Molarity, M = 0.143 M volume = V = 22 mL = 0.022 liter number of moles, n

M = n / V (in liter) ⇒ n - M * V = 0.143 M * 0.022 liter = 0.033146 mol

4) Number of moles of AgNO₃

Set a proportion, using the appropiate mole ratio and the number of moles of MgCl₂:

1 mol MgCl₂ / 2 mol AgNO₃ = 0.033146 mol MgCl₂ / x

Solve for x:

x = 0.033146 mol MgCl₂ * 2 mol AgNO₃ / 1 mol MgCl₂ = 0.006292 mol AgNO₃

4) Molarity AgNO₃

M = n / V (in liter) = 0.006292 mol / 15 mL * 1000 mL / liter = 0.41947

Round to two significant figures: 0.42 M ← answer