50.0 grams of RbNO3 (MM is 147.48 g/mol) and 75.0 grams of Rb were combined to react via the balanced equation shown below. How many grams of Rb2O (MM is 186.94 g/mol) should be formed?
10Rb + 2 RbNO3 -> 6Rb2O + 1N2
This is a limiting reagent (LR) problem. You know that when an amount is given for more than one of the reactants. I work LR problems the long way as follows.
10Rb + 2 RbNO3 -> 6Rb2O + 1N2
mols Rb = 75.0/85.5 = 0.877
mols RbNO3 = 50/147.5 = 0.338
mols Rb2O formed if all of the Rb were used. That's
0.877 mols Rb x (6 mols Rb2O/10 mols Rb) = 0.877 x 6/10 = 0.526
mols Rb2O formed if all of the RbNO3 were used. That's
0.338 mols RbNO3 x (6 mols Rb2O/2 mols RbNO3) = 0.338 x 6/2 = 1.02
In LR problems the small number is the winner since you can't form more moles than the least reactant so Rb is the LR.
mols Rb2O formed is mols Rb2O from Rb, which is 0.526 x molar mass Rb2O = grams Rb2O.
To find the grams of Rb2O formed, we need to use stoichiometry, which relates the amounts of different substances in a chemical reaction.
First, let's calculate the moles of RbNO3 and Rb using their given masses and molar masses.
Moles of RbNO3 = Mass of RbNO3 / Molar mass of RbNO3
= 50.0 g / 147.48 g/mol
= 0.339 moles
Moles of Rb = Mass of Rb / Molar mass of Rb
= 75.0 g / 85.47 g/mol
= 0.877 moles
According to the balanced equation, the mole ratio between RbNO3 and Rb2O is 2:6. This means that for every 2 moles of RbNO3, we will get 6 moles of Rb2O.
Now, let's use this mole ratio to calculate the number of moles of Rb2O formed.
Moles of Rb2O = (0.339 moles RbNO3) × (6 moles Rb2O/2 moles RbNO3)
= 0.339 moles × 3
= 1.017 moles
Finally, let's calculate the mass of Rb2O using its molar mass.
Mass of Rb2O = Moles of Rb2O × Molar mass of Rb2O
= 1.017 moles × 186.94 g/mol
= 190.33 g
Therefore, approximately 190.33 grams of Rb2O should be formed.