Respuesta :

 Obviously only one of the H's was replaced as the product showed one mole of water So the ratio is 1 to 1.Moles of NaOH used Molarity x volume = 0.1 x0.015(L) = 0.0015 moles 
Mass of acid used is 0.0015 x 204 = 0.306g 
moles MM

hopefully this helps