What is the concentration of NOBr, if the concentration of NO was measured to be 0.89 M, Br2 was measured to be 0.562 M, and the equilibrium constant, K, is 1.3 × 10-2?

Answers

Answer 1

Answer:

0.076M = [NOBr]

Explanation:

For the reaction:

2NO + Br₂ ⇄ 2 NOBr

The equilibirum constant, K, is defined as:

[tex]K = \frac{[NOBr]^2}{[NO]^2[Br_2]}[/tex](1)

Replacing the concentrations and the equilibrium value in (1):

[tex]1.3x10^{-2} = \frac{[NOBr]^2}{[0.89]^2[0.562]}[/tex]

5.79x10⁻³ = [NOBr]²

0.076M = [NOBr]

I hope it helps!


Related Questions

How much HCl is produced from the reaction of an excess of HSbCl4 with 3 moles H2S in the following reaction? HSbCl4 + H2S → Sb2S3 + HCl (Remember to balance the equation.)

Answers

Answer:

We will produce 8.0 moles of HCl , this is 291.7 grams HCl

Explanation:

Step 1: Data given

Number moles of H2S = 3.0 moles

Step 2: The balanced equation

2HSbCl4 + 3H2S → Sb2S3 + 8HCl

Step 3: Calculate moles HCl

For 2 moles HSbCl4 we need 3 moles H2S to produce 1mol Sb2S3 and 8 moles HCl

For 3.0 moles H2S we'll have 8.0 moles HCl

Step 4: Calculate mass HCl

Mass HCl = moles HCl * molar mass HCl

Mass HCl = 8.0 moles * 36.46 g/mol

Mass HCl = 291.7 grams

We will produce 8.0 moles of HCl , this is 291.7 grams HCl

Final answer:

From the balanced equation, it is determined that 2 moles of HCl are produced from 1 mole of HSbCl4. Therefore, 6 moles of HCl will be produced from the reaction of an excess of HSbCl4 with 3 moles of H2S.

Explanation:

The balanced equation for the reaction is:

HSbCl4 + H2S → Sb2S3 + 2HCl

The mole ratio between HSbCl4 and HCl is 1:2, which means that for every 1 mole of HSbCl4, 2 moles of HCl are produced.

Since there is an excess of HSbCl4, we can assume that all 3 moles of H2S will react.

Therefore, the number of moles of HCl produced will be:

(3 moles H2S) x (2 moles HCl/1 mole HSbCl4) = 6 moles HCl

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