Answer:
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
