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irakobra [83]
3 years ago
13

if the percent yield ofr the following reaction is 75% and 45.0g of NO2 are consumed with ercess water in the reaction. how many

grams of nitric acid HNO3(aq) are produced.
Chemistry
1 answer:
NISA [10]3 years ago
6 0

Answer:

30.8 grams of nitric acid are produced

Explanation:

Let's state the reaction:

3 NO₂ + H₂O → 2 HNO₃ + NO

If water is the excess reagent, then the limiting is the gas.

We convert the mass to moles:

45 g . 1 mol/ 46 g = 0.978 moles

Ratio is 3:2. 3 moles of gas can produce 2 moles of acid

Then, 0.978 moles may produce (0.978 . 2) /3 = 0.652 moles of acid

This is the 100% yield, but in this case, the percent yield is 75%

0.652 moles . 0.75 = 0.489 moles

Let's convert the moles to mass → 0.489 mol . 63g / 1mol = 30.8 g

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1. Calculate the average reaction rate expressed in moles H2 consumed per liter per second.
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Answer:

1) 0.0025 mol/L.s.

2) 0.0025 mol/L.s.

Explanation:

  • For the reaction:

<em>H₂ + Cl₂ → 2HCl.</em>

<em></em>

<em>The average reaction rate = - Δ[H₂]/Δt = - Δ[Cl₂]/Δt = 1/2 Δ[HCl]/Δt</em>

<em></em>

<em>1. Calculate the average reaction rate expressed in moles H₂ consumed per liter per second.</em>

<em></em>

The average reaction rate expressed in moles H₂ consumed per liter per second = - Δ[H₂]/Δt = - (0.02 M - 0.03 M)/(4.0 s) = 0.0025 mol/L.s.

<em>2. Calculate the average reaction rate expressed in moles CI₂ consumed per liter per second.</em>

<em></em>

The average reaction rate expressed in moles Cl₂ consumed per liter per second = - Δ[Cl₂]/Δt = - (0.04 M - 0.05 M)/(4.0 s) = 0.0025 mol/L.s.

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There are two main characters of matter:
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Mass and weight are the 2 main characters of matter
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Water pollution ________ salt water levels in the sea.
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A type of marble has a density of 2.5 grams/marble. A box filled with marbles has a total mass of 300 grams determine the number
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Answer:

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Explanation:

Step 1: Subtract 50 by 300 to find the total mass of the marbles

300 - 50 = 250

Not hard, right?

Step 2: Divide 250 by 2.5

250 ÷ 2.5 = 100

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Hope this helps :)

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