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tia_tia [17]
2 years ago
5

Consider the following reaction: Consider the reaction 2NO(g)+Br2(g)⇌2NOBr(g) ,Kp=28.4 at 298 K In a reaction mixture at equilib

rium, the partial pressure of NO is 119 torr and that of Br2 is 151 torr . What is the partial pressure of NOBr in this mixture?
Chemistry
1 answer:
Vitek1552 [10]2 years ago
4 0

Answer:  partial pressure of NOBr is 7792 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

2NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)

Equilibrium constant is given as:

K_{p}=\frac{[p_{NOBr}]^2}{[p_{NO}]^2\times [p_{Br_2}]^1}

28.4=\frac{[p_{NOBr}]^2}{[(119)^2\times (151)^1}

[p_{NOBr}]=7792 atm

Partial pressure of NOBr is 7792 atm

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Write a chemical equation of Dr. Jeff’s demonstration using a battery to light steel wool on fire. The battery is used to allow
Elis [28]

Answer:

2Fe(s) + 3O2(g) --------> 2FeO3(s)

Explanation:

According to the question, a  battery was used  to light the steel wool by bringing the terminals very close together. When  the battery came into contact with the steel wool, current was sent out through the thin wire.  This caused the iron to heat up quite well.

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This is the chemical reaction that occurs when the steel wool is set on fire.

6 0
2 years ago
What is the limiting reactant if 0.5 g Al is reacted with 3.5 g CuCl2?
DIA [1.3K]
The balanced equation for the above reaction is 
2Al + 3CuCl₂ --> 2AlCl₃ + 3Cu
stoichiometry of Al to CuCl₂ is 2:3
limiting reactant is when the reactant is fully consumed in the reaction therefore amount of product formed depends on amount of limiting reactant present. 
number of Al moles  - 0.5 g  / 27 g/mol = 0.019 mol
number of CuCl₂ moles - 3.5 g / 134.5 g/mol = 0.026 mol
if Al is the limiting reactant 
if 2 mol of Al reacts with 3 mol of CuCl₂
then 0.019 mol of Al reacts with - 3/2 x 0.019 = 0.029 mol of CuCl₂
but only 0.026 mol of CuCl₂ is present 
therefore CuCl₂ is the limiting reactant 
and 0.026 mol of CuCl₂ reacts with - 0.026/3 x 2 = 0.017 mol of Al is required
but 0.019 mol of Al is present 
therefore CuCl₂ is the limiting reactant and Al is in excess



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