When 3.648 grams of Epsom salt is heated to a constant mass, 1.782 grams of anhy

Question

When 3.648 grams of Epsom salt is heated to a constant mass, 1.782 grams of anhydrous magnesium sulfate is produced, according to the equation below. What is the formula for hydrated magnesium sulfate, also known as Epsom salt? What is the atom economy for the production of anhydrous magnesium sulfate from Epsom salt?MgSO4 x H2O rightarrow MgSO4 + x H2O Calculate the theoretical mass of sodium carbonate that should be produced from the decomposition of 1.800 grams of sodium bicarbonate. If 1.025 grams of sodium carbonate were generated from 1.800 grams of sodium bicarbonate, what is the percent yield of the reaction? A crucible containing 1.00 gram of sodium bicarbonate and 1.00 gram of sodium carbonate is heated to a constant mass. What mass of sodium carbonate remains in the crucible at the end of heating?

Explanation / Answer

Q. 1)

Given mass of Epsom salt = 3.648

Unhydrous magenesium sulfate = 1.782 g

Calculation of mass of H2O

= 3.648 g - 1.782 g =1.866 g

Moles of H2O

= 1.866 g / 18.0148 g per mol

= 0.1036 mol H2O

Moles of MgSO4 = 1.782 g / 120.366 g per mol

= 0.0148 mol

Mol ratio

Moles of H2O = 0.1036 / 0.0148 = 7

Moles of MgSO4 = 1

Hydrated formula = MgSO4.7H2O

Atom economy

= (Mass of desired product/ Total mass of all reactant ) *100

Lets calculate mass of desired product

Moles of reactant = 3.648 g / molar mass of MgSO4.7H2O

= 3.648 g/ 246.52 g per mol

= 0.0148 mol MgSO4 . 7 H2O

Moles of MgSO4

= 0.0148 mol MgSO4 .7H2O x 1 mol MgSO4/ 1 mol epsom

= 0.0148 mol MgSO4

Mass of MgSO4 = 0.0148 mol x 120.366 g /mol

=1.78 g

Atom economy = 1.78 / 3.648 x 100 = 49.0 %

2.)

Decomposition reaction

2 NaHCO3 -- > Na2CO3 + CO2 + H2O (g)

Calculation of moles of sodium bicarbonate

Moles of sodium bicarbonate = Mass / molar mass

= 1.800 g sodium bicarbonate / 84.007 g per mol

=0.02143 mol

Moles of sodium carbonate

= 0.02143 mol NaHCO3 x 1 mol Na2CO3/ 1 mol NaHCO3

=0.0107 mol

Mass of sodium carbonate = 0.0107 mol x Molar mass Na2CO3

=0.0107*106.0 g per mol

=1.14 g

Theoretical yield of sodium carbonate = 1.14 g

3).

Reaction

2 NaHCO3 -- > Na2CO3 + CO2 + H2O (g)

We got theoretical yield of Na2CO3 in problem 2.

Theoretical yield = 1.14

Actual yield = 1.025 g

Percent yield = actual yield / theoretical yield x 100

= 1.025 /1.14 x 100

= 90.0 %

4).

Mass of sodium carbonate = 1.00 g

Mass of sodium bicarbate = 1.00 g

Moles of sodium bicarbonate = 1.00 g / 84.007 g per mol

=0.0119 mol

Moles of sodium carbonate

=0.0119 NaHCO3 x 1mol Na2CO3/2 sodium bicarbonate

=0.00595 mol Na2CO3

Mass of Na2CO3 = 0.00595 mol x 106.0 g per mol

= 0.631 g

Total mass of sodium carbonate = 1.0 g + 0.631 g = 1.631 g

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