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3 years ago

The combustion of propane (C3H8) in the presence of excess oxygen yields CO2 and H2O:

Chemistry

1 answer:

White raven [17]3 years ago

4 0

Answer:

1.5 moles

Explanation:

Given data:

Number of moles of O₂ consumed = 2.5 mol

Moles of CO₂ produced = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Now we will compare the moles of oxygen with carbon dioxide from balance chemical equation:

O₂ : CO₂

5 : 3

2.5 : 3/5×2.5 = 1.5

Thus from 2.5 moles of oxygen 1.5 moles of carbon dioxide are produced.

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What is the rarefraction of a longitude wave

kvv77 [185]

Answer:

it is a transverse wave. Does that help?

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2 years ago

Nitrogen gas reacts with hydrogen gas to produce ammonia. How many liters of hydrogen gas at 95kPa and 15∘C are required to prod

viktelen [127]

Answer:

222.30 L

Explanation:

We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:

Mass of NH₃ = 100 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 100 / 17

Mole of NH₃ = 5.88 moles

Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e

Xmol of H₂ = (3 × 5.88)/2

Xmol of H₂ = 8.82 moles

Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:

Pressure (P) = 95 KPa

Temperature (T) = 15 °C = 15 + 273 = 288 K

Number of mole of H₂ (n) = 8.82 moles

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =?

PV = nRT

95 × V = 8.82 × 8.314 × 288

95 × V = 21118.89024

Divide both side by 95

V = 21118.89024 / 95

V = 222.30 L

Thus the volume of Hydrogen needed for the reaction is 222.30 L

8 0

2 years ago

2. As NH4OH is added to an HCl solution, the pH of the solution

chubhunter [2.5K]

Answer:

Explanation:

Nh4OH + HCL ---> NH4Cl + H3O

so ph decreases as H3O increases

and OH also decreases

5 0

2 years ago

Read 2 more answers

Who is interested in a middle schooler?

belka [17]

Answer:

I think that middle school teachers are interested in teaching middle schoolers.

6 0

2 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

6 0

3 years ago