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Brut [27]
3 years ago
14

Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is exce

ss NO2 present. The molar masses are as follows: H2O = 18.02 g/mol, HNO3 = 63.02 g/mol.
Chemistry
1 answer:
Iteru [2.4K]3 years ago
5 0

Answer:

10.85 g of water

Explanation:

First we write the balanced chemical equation

3NO_{2} +H_{2}O -->2HNO_{3} +NO

Then we calculate the number of moles of nitric acid produced

n(HNO3) = \frac{mass}{molar mass} =\frac{75.9g}{63.02g/mol}=1.2044 mol

According to the balanced equation, water needed in moles is always half the number of moles of HNO3 produced. So since we will produce 1.2044 mol of HNO3, we will need 0.6022 mol of water. Now to calculate what mass that is:

mass(water)=number of moles*molar mass=0.6022mol*18.02g/mol=10.85g

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Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
8 0
3 years ago
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