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.
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
1 answer
1 answer