Acetone is treated with excess of ethanol in the presence of hydrochloric acid. The product obtained is:

In the presence of dry HCl, one molecule of ethanol adds to the carbonyl group of acetone to give a hemiketal, $(CH_3)_2C(OH)(OC_2H_5)$.
With excess ethanol the hemiketal reacts with a second molecule of ethanol, losing water, to give the ketal.
$(CH_3)_2C=O + 2C_2H_5OH \xrightarrow{HCl} (CH_3)_2C(OC_2H_5)_2 + H_2O$
So the product is 2,2-diethoxypropane, $(CH_3)_2C(OC_2H_5)_2$.