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

the answer is b & c i believe : )