Propanone (1 mol) is treated with benzaldehyde (2 mol) in aqueous ethanolic $\mathrm{NaOH}$ at room temperature, and both aldol additions go on to dehydration. The product is

Benzaldehyde has no alpha H, so only propanone forms the enolate. Each of its two $\mathrm{CH}{}_3$ groups adds to one $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{CHO}$, and each aldol loses water to give the conjugated $\mathrm{C}{}_6\mathrm{H}{}_5-\mathrm{CH}=\mathrm{CH}-\mathrm{CO}-\mathrm{CH}=\mathrm{CH}-\mathrm{C}{}_6\mathrm{H}{}_5$. The 1:1 products (4-hydroxy-4-phenylbutan-2-one and 4-phenylbut-3-en-2-one) are intermediates, not the final product with 2 mol of aldehyde.