The water of 17.99 g forms when 63g of HNO₃ mixes with60g of NaOH
When nitric acid and sodium hydroxide are mixed, a neutralization reaction takes place, with the salt created by the two substances and water serving as the reaction result. The chemical reaction to the statement in this instance is:
HNO₃ + Na(OH) ⇒ NaNO₃ + H₂O
It is determined whether there is one limiting reagent and one in excess, or if both react entirely, by calculating the molar masses of the reagents and using the masses provided by the statement to compute the ratio in which these compounds are combined.
• The formula for HNO₃'s molar mass is 3xmO + mH + mN, or 3x15.99g + 1g + 14g = 62.97 g/mol.
• The formula for Na(OHmolar )'s mass is: mO + mH + mNa = 15.99g + 1.00g + 22.99g = 39.98 g / mol.
Nitric acid is immediately apparent to be the limiting reagent.
Since there are initially 60 g of hydroxide, some of it will stay unreacted because the balanced equation's stoichiometric ratio states that 1 mole of acid reacts with 1 mole of hydroxide. As a result, 63 g of HNO3 react with 40 g of hydroxide. This proves that acid is the limiting reagent, and calculations to determine how much water will form must be based on this substance.
For 63g of HNO₃ 18g of water is formed. Then,
for 62.97 g of HNO₃,
= 18 g x 63 g equals/62.97 g
= 17.99 g of water.
As a result, the correct response is that 17.99 g of water will form.
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