C) What mass of magnesium fluoride forms when 1.32 L of fluorine reacts? D) How

Question

C) What mass of magnesium fluoride forms when 1.32 L of fluorine reacts? D) How many formula units of magnesium fluoride can be produced by the reaction of 4.32 x 102 molecules of fluorine? 4. Calcium chloride reacts with sodium sulfate to form solid calcium sulfate precipitate CaCl + Na,so, a Caso, 2Naci SO, à CaSo, 2NaCI A) If 78.0 grams of Caci, and 76.6 grams of sodium sulfate are available. which reactant is the limiting reagent? Which is the excess reagent? What is the maximum mass (theoretical yield) of sodium chloride that can be produced in this reaction?

Explanation / Answer

CaCl2 + Na2SO4 ------------ CaSO4 + 2 NaCl

mass of CaCl2= 78.0grams

molar mass of CaCl2= 111 gras/mole

number of moles of CaCl2= 78.0/111 = 0.703 moles

mass of Na2SO4 = 76.6 grams

molar mass of Na2SO4 = 142.04 gram/mole

number of moles of Na2SO4 = 76.6/142.04 = 0.539 moles

According to equation

1 mole of CaCl2 = 1 mole of Na2SO4

number of moles of Na2SO4 are less compare the number of moles of CaCl2.

i,e Na2SO4 is consumed first. Hence Na2SO4 is limiting reagent.

CaCl2 is excess reagent,

According to equation

1 mole of Na2SO4 = 2 moles of NaCl

0.539 moles of Na2SO4 =?

                                     = 2 x 0.539/1 = 1.078 moles of NaCl

number of moles of NaCl= 1.078 moles

molar mass of NaCl = 58.5 gram/mole

mass of 1.078 moles of NaCl= 1.078 x 58.5 = 63.063 grams

Theoritical yield of NaCl = 63.06 grams.

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