Calculate the mass of MgCO3 precipitated by mixing 10.00 mL of a 0.200 M Na2CO3 solution with 5.00 mL of a 0.0450 M Mg (NO3) 2 solution.

1.90 g

Explanation:

When we are making calculations based on chemical reactions, we need first to have a balanced chemical equation.

From there we can determine the mass of the product MgCO₃ in this question.

Na₂CO₃ + Mg (NO₃) ₂ ⇒ 2 NaNO₃ + MgCO₃ (double decomposition)

0.200 M 0.0450 M?

10.0 5.00 mL

Now we know the volume and concentration of Mg (NO₃) ₂, and Na₂CO₃ so we must compute their number of moles to determine the limiting reagent, if any. From there determine moles and mass of MgCO₃ produced.

First lets convert the volume of Mg (NO₃) ₂and Na₂CO₃ to liters:

5.00 mL x (1 L/1000 mL) = 5.00 x 10⁻³ L

10.00 mL x (1L / 1000 mL) = 1.000 x 10 ⁻² L

# mol Mg (NO₃) ₂ = (0.0450 mol / L) x 5.00 x 10⁻³ L

= 2.25 x 10⁻⁴ mol Mg (NO₃) ₂

# mol Na₂CO₃ = (0.200 mol / L) x 1 x 10⁻² L

= 2.000 x 10⁻³ mol Na₂CO₃

Calculation limiting reagent:

= 2.25 x 10⁻⁴ mol Mg (NO₃) ₂ x (1 mol Na₂CO₃ / mol Mg (NO₃) ₂)

= 2.25 x 10⁻⁴ mol Na₂CO₃ required to react

Therefore, our limiting reagent is Mg (NO₃) ₂ since we require 2.25 x 10⁻⁴ mol Na₂CO₃ to react completely with 2.25 x 10⁻⁴Mg (NO₃) ₂, and we have excess of it.

# mol MgCO₃ produced

= 2.25 x 10⁻⁴ mol Mg (NO₃) ₂ x (1 mol MgCO₃ / 1 mol Mg (NO₃) ₂)

= 2.25 x 10⁻⁴ mol MgCO₃

Now that we have the moles of MgCO₃, we can obtain its mass by multiplying its molar mas (84.31 g/mol):

2.25 x 10⁻⁴ mol MgCO₃ x 84.31 g/mol = 1.90 g