If you had Pr6O11 + HNO3 and want to form Pr(NO3)3 * 6H2O, how would you figure out the number of grams of Pr6O11 and HNO3 you need?

Question

If you had Pr6O11 + HNO3 and want to form Pr(NO3)3 * 6H2O, how would you figure out the number of grams of Pr6O11 and HNO3 you need to form the Pr Nitrate Hexahydrate?

Assume the HNO3 is 68%.

Pr6O11: 1021.44 g/mol; HNO3: 63.01 g/mol



Need help finding the proper Balanced Equation and how to find the grams of each compound needed.

Answer 1

Pr₆O₁₁ is one form of praseodymium oxide.

Pr₆O₁₁ + 18HNO₃ + 27H₂O ==> 6Pr(NO₃)₃·6H₂O + O₂

I think this is the properly balanced equation obtained by balancing individual redox half reactions.

According to this equation you would want to use 1 mole of P₆O₁₁ and 18 moles of HNO₃.

1 mol P₆O₁₁ = 1021.44 g

18 moles HNO₃: 18 moles x 63.01 g / mol = 1134 g HNO₃

Since it is 68% (m/m) then 1134 g / 0.68 = 1668 g of the HNO₃ solution

So, in summary, assuming the balanced equation is the correct one (I believe it is)...

React 1021 g of P₆O₁₁ with 1668 g of the 68% HNO₃ in an aqueous medium, and you should produce 6Pr(NO₃)₃·6H₂O