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

Solution :-

Q1) Mass of hydrate = 3.648 g

Mass of anhydrate = 1.782 g

Mass of water lost = 3.648 g – 1.782 g = 1.866 g

Now lets calculate moles of the water and CuSO4

Moles of water = 1.866 g / 18.0148 g per mol = 0.103581 mol

Moles of CuSO4 = 1.782 g / 159.609 g per mol = 0.011165 mol

Now lets find the ratio of the water to CuSO4

Moles of water /moles of CuSO4 = ?

0.103581 mol / 0.011165 mol = 9.3 rounded to 9

So the formula of the hydrate = CuSO4. 9H2O

Percent atom economy = (1.782 g / 3.648 g) * 100% = 49 %

Q2)

Balanced reaction equation is as follows

2 NaHCO3 --------> Na2CO3 + CO2 + H2O

1.800 g NaHCO3

Na2CO3 = ? gram

Lets calculate the mass of the Na2CO3 using the mole ratio

(1.800 g NaHCO3 * 1 mol / 84.007 g)*(1 mol Na2CO3 / 2 mol NaHCO3) *(105.989g/ 1 mol Na2CO3) =1.136 g Na2CO3

Therefore the theoretical yield of the Na2CO3 = 1.136 g Na2CO3

Q3) % yield = (actual yield / theoretical yield) * 100 %

                      = (1.025 g / 1.136 g)*100%

                      = 90.23 %

Q4)

After the heating sodium bicarbonate will decompose to form the the sodium carbonate

So lets calculate the mass of the sodium carbonate that can be formed from the 1.00 g sodium bicarbonate

(1.0 g NaHCO3 * 1 mol / 84.007 g)*(1 mol Na2CO3 / 2 mol NaHCO3) *(105.989g/ 1 mol Na2CO3) =0.631 g Na2CO3

So the total mass of the carbonate in the crucible is 1.00 g + 0.631 g = 1.631 g

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