Sulfur trioxide is made from the oxidation of SO_2. and the reaction is represen

Question

Sulfur trioxide is made from the oxidation of SO_2. and the reaction is represented by the equation: 2SO_2 + O2 right arrow 2SO_3 A 21 -g sample of SO_2 gives 18 g of SO_3. The percent yield of SO_3 is The commercial production of phosphoric acid. H_3PO_4. Can be represented by the equation 1500 g 300 g 307 g 1180 g 300 g Ca_3 (PO_4)_2 + 3SiO_2 + 5C + 5O_2 + 3H_2O right arrow 3CaSiO_3 + 5CO_2 + 2H_3PO_4 310 g/mol 60.1 g/mol 12.0 g/mol 32.0 g/mol 18.0 g/mol The molar mass for each reactant is shown below the reactant, and the mass of each reactant for this problem is given above. Which substance is the limiting reactant?

Explanation / Answer

11) Molar mass of SO2 = 32+2x16 = 64 No. of moles of SO2= 21/64 = 0.3281

Since it is not provided in question i am assuming 1 mole of O2 is used.

For every 1 mole of oxygen 2 moles of SO2is required we have 0.3281 moles of SO2 so we need 0.5 moles of O2

But we have 1 mole of oxygen, so O2 is the excess reagent and SO2 is the limiting reagent.

Since 0.3281 moles of SO2 is needed, it produces 0.3281 moles of SO3

Mass of SO3 produced theoretically = 0.3281 x 80 = 26.248 g

Actual yield = 18 g

% yield = (Actual yield / theoretical yield) x 100 = (18/26.248) x 100 = 68.57 %

12) No. of moles of Ca3(PO4)2 = 1500/310 = 4.838

No. of moles of SiO2 = 300/60.1 = 4.991

No of moles of Carbon = 307/12 = 25.583

   No . of moles of O2 = 1180/32 = 36.875

No. of moles of water = 300 / 18 =16.66

SiO2 = 4.991/3 = 1.663

Ca3(PO4)2 = 4.838

Clearly SiO2 is the limiting reagent.

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