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Sindrei [870]
1 year ago
13

Chromium (III) oxide reacts with hydrogen sulfide gas to form chromium (III) sulfide and water. To produce 421 g of cr2s3, how m

any moles of cr2o3 and grams of cr2o3 are required?
Chemistry
1 answer:
maw [93]1 year ago
6 0

Cr₂O₃ ( s ) + 3H₂S ( g ) → Cr₂S₃ ( s ) + 3H₂O ( l )

mol  Cr₂S₃ = 421 : 200.19 g/mol = 2.103

mol Cr₂O₃ ≈ mol Cr₂S₃ = 2.103 ( equivalent coefficient)

mass Cr₂O₃ = 2.103 x 151.99 g/mol = 319.63 gr

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If the initial [NO2] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M. If the initial is 0.260 , it
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The question is incomplete, here is the complete question:

At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas

NO_2\rightarrow NO+\frac{1}{2}O_2

The reaction is second order for NO_2 with a rate constant of 0.543M^{-1}s^{-1} at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M

a) 1.01    b) 5.19     c) 0.299      d) 0.0880     e) 3.34

<u>Answer:</u> The time taken is 5.19 seconds

<u>Explanation:</u>

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.543M^{-1}s^{-1}

t = time taken  = ?

[A] = concentration of substance after time 't' = 0.150 M

[A]_o = Initial concentration = 0.260 M

Putting values in above equation, we get:

0.543=\frac{1}{t}\left (\frac{1}{(0.150)}-\frac{1}{(0.260)}\right)\\\\t=5.19s

Hence, the time taken is 5.19 seconds

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