Calculate the volume of 0.07216 M AgNO3 needed to react exactly with 0.3572 g of pure Na2CO3 to produce solid Ag2CO3.

93.4 mL

Explanation:

Let's state the reaction:

2AgNO₃ +  Na₂CO₃  →   Ag₂CO₃  +  2NaNO₃

We determine the moles of sodium carbonate:

0.3572 g . 1mol / 105.98g = 3.37×10⁻³ moles

Ratio is 1:2. We say:

1 mol of sodium carbonate react to 2 moles of silver nitrate

Then, our 3.37×10⁻³ moles of carbonate may react to: 3.37×10⁻³ . 2

= 6.74×10⁻³ moles

If we convert to mmoles → 6.74×10⁻³ mol . 1000 mmol / mol = 6.74 mmol

Molarity is mol/L but we can use mmol /mL

6.74 mol / volume in mL = 0.07216 M

6.74 mol / 0.07216 M = volume in mL → 93.4 mL

14,16 18 reacts with group

it's c. elements are placed in rows based on how may valencies electrons have.