23. Given the following reaction reaction CaSiO3 + 6HF + CaF2 + SiF4 +3H2O If 25

Question

23. Given the following reaction reaction
CaSiO3 + 6HF + CaF2 + SiF4 +3H2O
If 250.0g of CaSiO3 were reacted with 250.0g of HF
Give your answers in the correct number of significant digits
(a) What is the limiting Reagent?

(b) How many grams of water will be produced?
(c) What is the mass of the excess reagent remaining after the reaction has occurred?

(d) If 50.0g of water is produced what is the yield of the reaction
(e) If the reaction yield is 76.2% what mass of water will be produced

(f) How many grams of HF must be used to produce 540.0gof CaF2
(g) If the yield of the reaction is 76.2% how many grams of HF must be used to
produce 540.0g of CaF2 .

Explanation / Answer

a-CaSiO3

b-mw ofCaSiO3 = 116.16g/mol 250g =2.1522mol

mwof HF =20.g/mol 250g = 12.5mol

CaSiO3 + 6HF + CaF2 + SiF4 +3H2O

1mol            6mol                         3mol

12.5 mol of HF xx (3/6) x (18.0153 g H2O/mol) =112.596gH2O

c-6mol HF react with 1 mol CaSiO3

12.5 mol HF                12.5/6=2.08mol CaSiO3

amount of CaSiO3 left = 2.1522-2.08 mol= 0.0722mol=8.3867g

250-240= 10g left

d- % yield = 50x100/112.596 =44.406%

e-mass of water = 112.596 x76.2/100= 85.79g

f- mw of CaF2=78.07g/mol

mole of HF x1/6 x 78.07 = 540g

mole of HF = 540x6/78.07= 41.5mol =830.0g

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