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!
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.)
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
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