Suppose the reaction ca3(po4)2 + 3h2so4 ® 3caso4 + 2h3po4 is carried out starting with 153 g of ca3(po4)2 and 76.8 g of h2so4. how much phosphoric acid will be produced?

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Solution:- The given balanced equation is:

From this equation, there is 1:2 mol ratio between calcium phosphate and phosphiric acid. The mol ratio between sulfuric acid and phosphoric acid is 3:2.

First of all we convert the grams of each reactant into moles and then the moles of each are multiplied by the above mol ratios to get the moles of phosphoric acid. Moles of phosphoric acid are multiplied by it's molar mass to get it's grams. The reactant that gives less grams of phosphoric acid would be the limiting reactant and this amount of phosphoric acid formed will be our answer.

The calculations for grams of phosphoric acid calculated from each of the reactants are as follows:

Calculations of phosphoric acid grams from given grams of calcium phosphate:

=

Calculations for grams of phosphoric acid from given grams of sulfuric acid:

=

From above calculations, sulfuric acid gives least amount of phosphoric acid. It means sulfuric acid is limiting reactant and the amount of phosphoric acid formed is 51.2 g.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally