Question

Mixing solutions of sodium hydroxide and rubidium nitrate will cause rubidium hydroxide to precipitate. You have a 1.50 mol/L sodium hydroxide solution and a 0.200 mol/L rubidium nitration solution. What volume of each solution should be mixed together to form 15.0 g of rubidium hydroxide?

Answer 1

97.6 mL of Sodium Hydroxide and 732 mL of Rubidium Nitrate are needed to make 15 g of Rubidium hydroxide.

Step-by-step explanation:

The chemical reaction described in the question can be given as

NaOH + RbNO₃ → RbOH + NaNO₃

We want to find the amount of each reactant that reacted to give 15 g of RbOH

Amount of RbOH produced from the reaction = 15 g,

We first convert to number of moles, to be able to obtain the number of moles of each reactant from the stoichiometric balance of the reaction.

Number of moles = (Mass)/(Molar Mass)

Molar Mass of RbOH = 102.475 g/mol

Number of moles of RbOH produced from the reaction = (15/102.475) = 0.1464 moles.

Note that from the stoichiometric balance of the reaction,

1 mole of RbOH was produced from 1 mole of RbNO₃

And

1 mole of RbOH was produced from 1 mole of NaOH.

So,

0.1464 mole of RbOH will be produced from 0.1464 mole of RbNO₃

And

0.1464 mole of RbOH will be produced from 0.1464 mole of NaOH

Note that,

Number of moles = (Concentration in mol/L) × (Volume in L)

For RbNO₃,

Number of moles = 0.1464 mole

Concentration in mol/L = 0.200 mol/L

Volume in L = ?

0.1464 = 0.2 × (Volume in L)

Volume in L = (0.1464/0.20) = 0.732 L = 732 mL

For NaOH,

Number of moles = 0.1464 mole

Concentration in mol/L = 1.50 mol/L

Volume in L = ?

0.1464 = 1.50 × (Volume in L)

Volume in L = (0.1464/1.50) = 0.0976 L = 97.6 mL

Hope this Helps!!!