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3 years ago

Identify the limiting reagent and the volume of CO2 formed when 11L Cs2 reacts with 18L O2 to produce CO2 gas?

Chemistry

2 answers:

frozen [14]3 years ago

7 0

Answer:

The correct answer is option a) O2; 6.0 L C02

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Then, taking into account that the reaction occurs under STP conditions, it is possible to determine the amount of moles of CS2 and O2 that react, with volumes 11L and 18L respectively. This determination is made by a rule of three:

if 1 mole of gas occupies 22.4 L, how many moles of CS2 occupy 11 L?

$moles of CS2=\frac{11 L*1 moles}{0.49 L}$

moles of CS2=0.49

if 1 mole of gas occupies 22.4 L, how many moles of O2 occupy 18 L?

$moles of O2=\frac{18 L* 1 moles}{22.4 L}$

moles of O2=0.80

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), and the amount of moles calculated previously. You can use a simple rule of three as follows: if by stoichiometry one mole of CS2 reacts with 3 moles of O2, how much moles of O2 will be needed if 0.49 moles of CS2 react?

$moles of O2=\frac{0.49 moles of CS2*3moles of O2}{1 mol of CS2}$

moles of O2=1.47

But 1.47 moles of O2 are not available, 0.80 moles are available. Since you have less moles than you need to react with 0.49 moles of CS2, oxygen O2 will be the limiting reagent.

Then, it is possible to determine the amount of moles of CO2 produced by another rule of three: if by stoichiometry 3 moles of O2 produce 1 mole of CO2, how many moles of CO2 will be formed if 0.80 moles of O2 react?

$moles of CO2=\frac{0.8 moles of O2*1 mol de CO2}{3 moles of O2}$

moles of CO2=0.27

Finally, taking into account the previously stated STP conditions and the amount of moles of CO2 formed, it is possible to determine the volume of CO2 that is formed by a rule of three: if 1 mole occupies 22.4 L, how much volume occupies 0.27 moles?

$Volume of CO2=\frac{0.27 moles* 22.4 L}{1 mol}$

Volume of CO2=6.048 L ≅ 6.0 L

So, the correct answer is option a) O2; 6.0 L C02

lidiya [134]3 years ago

5 0

He limiting reagent (or limiting reactant) in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. We determine by doing as follows:

11 L Cs2 ( 1 mol / 22.4 L ) = 0.49 mol Cs2
18 L O2 ( 1 mol / 22.4 L ) = 0.80 mol O2

Therefore, the limiting reactant would be oxygen since it only needs about 0.27 mol of Cs2 to be consumed completely.

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a gas expands from 266 mL to 983 mL at constant temperature. calc the work done (in joules) by the gas if it expands against a v

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Answer:

w₁ = 0 J

w₂ = -297 J

Explanation:

Step 1: Given data

Initial volume: 266 mL (0.266 L)

Final volume: 983 mL (0.983 L)

Step 2: Calculate the work done (in joules) by the gas if it expands against a vacuum

We will use the following expression.

w₁ = -P₁ × ΔV

Since the gas expands against a vacuum, P₁ = 0. Thus, w₁ = 0 J

Step 3: Calculate the work done (in joules) by the gas if it expands against a constant pressure of 4.09 atm

We will use the following expression.

w₂ = -P₂ × ΔV

w₂ = -4.09 atm × (0.983 L-0.266 L) = -2.93 atm.L

Then, we convert w₂ to Joule using the conversion factor 1 atm.L = 101.325 J.

-2.93 atm.L × 101.325 J/1 atm.L = -297 J

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