Nitric acid (63 g) and sodium hydroxide (60 g) are mixed. how many grams of water will form.?

17.99 g of water

Explanation:

When nitric acid is combined with sodium hydroxide, a neutralization reaction occurs, which has as reaction products the salt formed by both compounds and water. In the case of the statement, the balanced reaction is:

HNO₃ + Na(OH) ⇒ NaNO₃ + H₂O

The molar masses of the reagents are calculated, and with the masses provided by the statement, the ratio in which these compounds are combined is calculated and it is determined whether there is one limiting reagent and one in excess, or if both react completely.

• Molar mass of HNO₃ = 3xmO + mH + mN = 3x15.99g + 1.00g + 14.00g = 62.97 g / mol

• Molar mass of Na (OH) = mO + mH + mNa = 15.99g + 1.00g + 22.99g = 39.98 g / mol

It can be seen at a glance that nitric acid is the limiting reagent, because the stoichiometric ratio in the balanced equation is that 1 mole of acid reacts with one mole of hydroxide, therefore 63 g of HNO₃ react with 40 g of hydroxide, and Since initially there is 60 g of hydroxide, a part of it will remain unreacted. This establishes that the limiting reagent is acid, and calculations to know how much water will be formed must be done based on this compound.

63g of HNO₃ 18g of water

62.97 g of HNO₃ X = 17.99 g of water

Calculation: 62.97 g x 18 g / 63g = 17.99 g of water

Therefore, the answer is that 17.99 g of water will form.

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.