The mass of MgCO₃ is 1.9g.
The balanced chemical reaction is shown below
Na₂CO₃ + Mg(NO₃)₂ ⇒ 2 NaNO₃ + MgCO₃
0.200 M 0.0450 M ?
10.0 5.00 mL ?
Since the volume and concentration of Mg(NO₃)₂ and Na₂CO₃ is given , we can calculate the number of moles for each of them and then determine the limiting reagent.
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
Number of mol Mg(NO₃)₂ = ( 0.0450 mol /L ) x 5.00 x 10⁻³ L
= 2.25 x 10⁻⁴ mol Mg(NO₃)₂
Number of mol Na₂CO₃ = ( 0.200 mol / L ) x 1 x 10⁻² L
= 2.000 x 10⁻³ mol Na₂CO₃
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 .
Limiting reagent is Mg(NO₃)₂ since 2.25 x 10⁻⁴ mol Na₂CO₃ is required to
react completely with 2.25 x 10⁻⁴Mg(NO₃)₂, and there's an excess.
Number of mole of 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
Mole = mass/molar mass
Mass= Mole × molar mass
2.25 x 10⁻⁴ mol MgCO₃ x 84.31 g/mol = 1.90 g
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