A solution containing 0.1 mol of nitric acid was added to 10 g of potassium hydroxide

A solution containing 0.1 mol of nitric acid was added to 10 g of potassium hydroxide. What substance and in what quantity will remain in excess?

Nitric acid react with potassium hydroxide:
KOH + HNO3 = KNO3 + H2O.
Determine the number of mol of potassium hydroxide:
n (KOH) = m / M = 10/56 = 0.1786 mol.
From the condition of the problem, n (HNO3) = 0.1 mol.
From the reaction equation, it can be seen that an equal amount of moles of substances enters into the reaction, respectively, nitric acid will be in short supply, and alkali in excess:
n (KOH) = 0.1786 – 0.1 = 0.0786 mol.