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r-ruslan [8.4K]
4 years ago
9

g Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.71 g of ethane i

s mixed with 3.8 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Kamila [148]4 years ago
8 0

Answer:

6.05g

Explanation:

The reaction is given as;

Ethane + oxygen --> Carbon dioxide + water

2C2H6 + 7O2 --> 4CO2 + 6H2O

From the reaction above;

2 mol of ethane reacts with 7 mol of oxygen.

To proceed, we have to obtain the limiting reagent,

2,71g of ethane;

Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol

3.8g of oxygen;

Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol

If 0.0903 moles of ethane was used, it would require;

2 = 7

0.0903 = x

x = 0.31605 mol of oxygen needed

This means that oxygen is our limiting reagent.

From the reaction,

7 mol of oxygen yields 4 mol of carbon dioxide

0.2375 yields x?

7 = 4

0.2375 = x

x = 0.1357

Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g

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Explanation:

The full ionic chemical balanced equation for CuCl2 and (NH4)2SO4 is

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The right siders are products i.e.,(NH4)Cl and CuSO4.

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Calculate the total amount of energy required to change 10.0 g of water from 35.0 degrees Celsius to 110. degrees Celsius.
Makovka662 [10]

Answer:

The total amount of energy required is 25,515.2 J.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

When a system absorbs (or gives up) a certain amount of heat, it can happen that:

  • experience a change in its temperature, which involves sensible heat,
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To calculate the latent heat the formula is used:

Q = m. L

Where

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To calculate sensible heat the following formula is used:

Q = m. c. ΔT

where:

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In this case, you have in the first place a heat to raise the temp of the water from 35.0 C to 100 C, where the specific heat value for water is  4.184 \frac{J}{g*C}:

q1 = m*c*(Tfinal-Tinitial)

q1 = 10.0 g *(4.184 \frac{J}{g*C})* (100 - 35.0 C) = 2719.6 J

Now you have the heat to vaporize the water, where the heat of vaporization is 2259.36 \frac{J}{g}:

q2 = m*(heat of vaporization)

q2 = 10.0 g*(2259.36 \frac{J}{g}) = 22593.6 J

Finally, you have the heat to raise temp of steam to 110 C, where the specific heat value for steam is  2.02 \frac{J}{g*C} :

q3 = m*c*(Tfinal-Tinitial)

q3 = 10.0 g*(2.02 \frac{J}{g*C})*(110-100 C) = 202 J

The total amount of energy can be calculated as:

Q= q1 + q2 + q3

Q= 2719.6 J + 22593.6 J + 202 J

Q=25,515.2 J

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Explanation:

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