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3 years ago

Prob.: Consider the combustion of butane (C4H10):

Chemistry

1 answer:

never [62]3 years ago

7 0

<u>Answer:</u> The moles of carbon dioxide formed in the reaction is 20 moles.

<u>Explanation:</u>

We are given:

Number of moles of butane = 5.0 moles

The chemical reaction for the combustion of butane follows the equation:

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(l)$

As, oxygen is present in excess. So, it is considered as an excess reagent.

Thus, butane is considered as a limiting reagent because it limits the formation of products.

By stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide.

So, 5 moles of butane will produce = $\frac{8}{2}\times 5=20mol$ of carbon dioxide.

Hence, the moles of carbon dioxide formed in the reaction is 20 moles.

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lord [1]

Answer:

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Explanation:

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3 years ago

Which contains the greatest mass of oxygen: 0.75 moles of ethanol ( C2H5OH ), 0.60 mole of formic acid ( HCO2H ), or 1.0 mole of

Luda [366]

Answer:

In conclussion, 0.60 moles of HCOOH contains the greatest mass of O

Explanation:

Let's make some rules of three, to solve this problem:

1 mol of ethanol has 2 moles of C, 6 moles of H, and 1 mol of oxygen

Therefore, 0.75 moles of ethanol must have 0.75 mol of oyxgen

Let's convert the moles to mass → 0.75 mol . 16 g/ 1 mol = 12 g

1 mol of formic acid has 2 moles of H, 1 mol of C and 2 mol of oxygen

0.60 moles of formic acid must have (0.6 .2) / 1 = 1.2 mol of O

If we convert the amount to mass → 1.2 mol . 16 g/ 1mol = 19.2 g

1 mol of water has 1 mol of oyxgen

Therefore, we have 1 mol of oxygen with a mass of 16 g.

In conclussion, 0.60 moles of HCOOH contains the greatest mass of O

7 0

3 years ago

Read 2 more answers

If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give

tankabanditka [31]

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Given:

The nitrogen gas molecule with a temperature of 330 Kelvins is released from Earth's surface to travel upward.

To find:

The maximum height of a nitrogen molecule when released from the Earth's surface before coming to rest.

Solution:

The maximum height attained by nitrogen gas molecule = h
The temperature of nitrogen gas particle = T = 330 K

The average kinetic energy of the gas particles is given by:

$K.E=\frac{3}{2}K_bT\\\\K.E=\frac{3}{2}\times 1.38\times 10^{-23} J/K\times 330 K\\\\K.E=6.381\times 10^{-21} J$

The nitrogen molecule at its maximum height will have zero kinetic energy as all the kinetic energy will get converted into potential energy

The potential energy at height h = $P.E = 6.381\times 10^{-21} J$
Molar mass of nitrogen gas = 28.0134 g/mol
Mass of nitrogen gas molecule = m

$m= \frac{ 28.0134 g/mol}{6.022\times 10^{23} mol^{-1}}=4.652\times 10^{-23} g\\\\1g=0.001kg\\\\m=4.652\times 10^{-23}\times 0.001 kg\\\\=4.652\times 10^{-26} kg$

The acceleration due to gravity = g = 9.8 m/s^2
The maximum height attained by nitrogen gas molecule = h
The potential energy is given by:

$P.E=mgh$

$6.381\times 10^{-21} J=4.652\times 10^{-26} kg\times 9.8 m/s^2\times h\\\\h=\frac{6.381\times 10^{-21} J}{4.652\times 10^{-26} kg\times 9.8 m/s^2}\\\\h=13,996.6 m\\\\1 m = 0.001 km\\\\h=13,996.6 m=h=13,996.6\times 0.001 k m\\\\=13.9966 km \approx 14 km$