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hram777 [196]
3 years ago
11

When nitrogen dioxide gas dissolves in water, an aqueous solution containing dissolved nitric acid and nitrogen monoxide forms.

In terms of mass balance, 138.03 g of NO_2 completely reacts with 18.02 g of H_2O to form 126.04 g of HNO_3 and 30.01 g of NO. 3NO_2(g) + H_2O(l) rightarrow 2HNO_3(aq) + NO(aq) 1. When 359 g of NO_2 is used, how many grams of H_2O are consumed? 2. How many grams of HNO_3 are produced? 3. How many grams of NO are produced?
Chemistry
1 answer:
xeze [42]3 years ago
5 0

Answer:

1. Based on the given question, 138.03 grams of NO₂ is reacting completely with 18.02 grams of H2O. However, in case when 359 grams of NO₂ is used then the grams of water consumed in the reaction will be,  

= 359 × 18.02 / 138.03 = 46.87 grams of water.  

2. As mentioned in the given case 138.03 grams of NO₂ generates 126.04 grams of HNO₃. Therefore, 359 grams of NO₂ will produce,  

= 359 × 126.04 / 138.03  

= 327.81 grams of HNO₃.  

3. Based on the given question, 138.04 grams of NO₂ is generating 30.01 grams of NO. Therefore, 359 grams of NO₂ will generate,  

= 359 × 30.01 / 138.04

= 78.04 grams of NO.  

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Given ; Current ( I ) = 5. 68 A

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x moles of  e^{-} are required for the decomposition of 0.128 mole of Fe .

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<em>Moles that react:</em>

NaOH = 10mL = 0.010L * (0,240mol / L) = 0.0024 moles NaOH

CH₃COOH = 50.0mL = 0.050L * (0.120mol / L) = 0.0060 moles CH₃COOH

That means after the reaction you will have:

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CH₃COO⁻Na⁺: 0.0024 moles

in solution, you will have the mixture of a weak acid (Acetic acid), with its conjugate base (sodium acetate, CH₃COO⁻Na⁺). And pH of this buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

For Acetic buffer pKa = 4.76:

pH = 4.76 + log [CH₃COO⁻Na⁺] / [CH₃COOH]

<em>Where [] is molarity of each species or moles</em>

<em />

Replacing:

pH = 4.76 + log [0.0024 moles] / [0.0036 moles]

<h3>pH = 4.58</h3>

<em />

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