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enyata [817]
3 years ago
6

N2 + 3H2, -> 2NH3 If I have 50.6 L of N2 and excess H2, how many liters of NH3 can I produce?

Chemistry
1 answer:
Anastaziya [24]3 years ago
3 0

Answer:

C. 101.2 L

Explanation:

N2 + H2= NH3

Balancing it,

N2 + 3 H2 = 2.NH3

(1 mol) (3 mol) (2 mol)

which means

1 molecule of nitrogen reacts with 3 molecule of hydrogen to form ammonia.

Likewise,

50.6 l of nitrogen reacts with 50.6 × 3= 151.8 l of hydrogrn to form 50.6 × 2= 101.2 l of ammonia.

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6 0
3 years ago
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The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper
LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

t_{1/2}=\frac{1}{k\times a_0}

k = rate constant =?

a_0 = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

a) Initial concentration of XY = a_o=0.100 M

Rate constant of the reaction = k = 6.96\times 10^{-3} M^{-1} s^{-1}

Half life of the reaction is:

t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}

=1,436.78 s

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = 6.20\times 10^{-2} M

\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become 6.20\times 10^{-2} M.

e)  Initial concentration of SO_2Cl_2= 0.050 M

Final concentration after time t = x

t = 55.0 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0
3 years ago
Use this equation for the next question:
MA_775_DIABLO [31]
The answers is: 375 moles NaOH
6 0
3 years ago
What is the molarity of a solution that contains 0.220 moles KOH in 0.350 L of solution?
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Answer:

0.628 M.

Explanation:

In order to solve this problem we need to keep in mind the<em> definition of molarity</em>:

  • Molarity = moles / liters

We are given both the <em>number of moles and the volume of solution</em>, meaning we can now proceed to <u>calculate the molarity</u>:

  • Molarity = 0.220 mol / 0.350 L
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