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