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 Mvolume = V = 22 mL = 0.022 liternumber of moles, nM = 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

The answer is c) pyrite

The correct answer is "hydrogen".