You have 75.0 mL of a 2.50 M solution of Na2CrO4(aq). How many mL of a 1.74 M solution of AgNO3(aq) are needed to completely react both reactants?

215 mL of silver nitrate are needed.

Explanation:

Let's think the reaction:

Na₂CrO₄(aq) + 2AgNO₃(aq) → Ag₂CrO₄ (s) ↓ + 2NaNO₃(aq)

First of all, we determine the moles of chromate we have available.

Molarity = mol / volume(L)

We convert volume to L → 75mL . 1L / 1000mL = 0.075 L

Molarity . volume(L) = moles → 2.50 M . 0.075L = 0.1875 moles

Ratio is 1:2. For 1 mol of chromate I need the double of moles of nitrate.

So, for 0.1875 moles of chromate I would need (0.1875 . 2) = 0.375 moles of nitrate. Let's determine volume of solution

Molarity = mol/ volume(L) → volume (L) = mol / Molarity

volume (L) = 0.375mol / 1.74M → 0.215L

We convert the value to mL → 0.215L . 1000mL/ 1L = 215 mL

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