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

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explanation:

explanation:

two years ago, when picking flowers in the mountains back of yosemite valley, i found a book. it was blotted and storm-beaten; all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however; many of the inner pages were well preserved. and although all were more or less stained and torn, whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today; its granite pages have been torn and blurred by the same storms that wasted the castaway book.

two years ago, when picking flowers in the mountains back of yosemite valley; i found a book. it was blotted and storm-beaten, all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however, many of the inner pages were well preserved. and although all were more or less stained and torn; whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today, its granite pages have been torn and blurred by the same storms that wasted the castaway book.