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Sergeeva-Olga [200]

3 years ago

8

Given the following equation: 4 NH3 (g)5 O2 (g) >4 NO (g) + 6 H20 (I) How many moles of NH3 is required to react with 25.7 gr

ams of O2?

1 answer:

Sergeu [11.5K]3 years ago

5 0

Answer:

0.6425 moles of $NH_3$ is required to react with 25.7 grams of $O_2$ .

Explanation:

mas of oxygen gas = 25.7 g

moles of oxygengas = $\frac{25.7 g}{32 g/mol}=0.8031 mol$

$4NH_3 (g)+5O_2 (g) \rightarrow 4 NO(g) + 6H_20 (I)$

According to reaction given above, 5 moles of oxygen gas reacts with 4 moles of ammonia gas.

Then 0.8031 moles of oxygen gas will react with :

$\frac{4}{5}\times 0.8031 mol=0.6425 mol$ of ammonia gas

0.6425 moles of $NH_3$ is required to react with 25.7 grams of $O_2$ .

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The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-

fomenos

Answer:

The total pressure after one half is 6.375 atm.

Explanation:

The initial pressure of product is increases while the pressure of reactant would decrease.

Balanced chemical equation:

2N₂O → 2N₂ + O₂

The pressure of N₂O is 5.10 atm. The change in pressure would be,

N₂O = -2x

N₂ = +2x

O₂ = +x

The total pressure will be

P(total) = P(N₂O) + P(N₂) + P(O₂)

P(total) = ( 5.10 - 2x) + (2x) + (x)

P(total) = 5.10 + x

After one half life:

P(N₂O) = 1/2(5.10) = 5.10 - 2x

x = 5.10 - 1/2(5.10) /2

x = 5.10 - 0.5 (5.10) /2

x = 5.10 - 2.55 / 2

x = 2.55 /2 = 1.275 atm

Thus the total pressure will be,

P(total) = 5.10 + x

P(total) = 5.10 + 1.275

P(total) = 6.375 atm

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A, D, E

Explanation:

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Read 2 more answers

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin

erastovalidia [21]

Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

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Tasya [4]

A mineral is a naturally occurring inorganic element or compound having an orderly internal structure and characteristic chemical composition, crystal form, and physical properties. ... A rock is an aggregate of one or more minerals, or a body of undifferentiated mineral matter.

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