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jeyben [28]
3 years ago
7

A cylinder contains 3.1 L of oxygen at 300 K and 2.7 atm. The gas is heated, causing a piston in the cylinder to move outward. T

he heating causes the temperature to rise to 610 K and the volume of the cylinder to increase to 9.4 L.
How many moles of gas are in the cylinder?
Express your answer using two significant figures.
Chemistry
1 answer:
zubka84 [21]3 years ago
6 0

Answer: The moles of gas present in the cylinder is 0.34 moles.

Explanation:

Given: P_{1} = 2.7 atm,   V_{1} = 3.1 L,     T_{1} = 300 K

P_{2} = ?,      V_{2} = 9.4 L,       T_{2} = 610 K

Formula used to calculate the final temperature is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{2.7 atm \times 3.1 L}{300 K} = \frac{P_{2} \times 9.4 L}{610 K}\\P_{2} = \frac{5105.7}{2820} atm\\= 1.81 atm

Now, moles present upon heating the cylinder are as follows.

P_{2}V_{2} = n_{2}RT_{2}\\1.81 atm \times 9.4 L = n_{2} \times 0.0821 L atm/mol K \times 610 K\\n_{2} = \frac{17.014}{50.081} mol\\= 0.34 mol

Thus, we can conclude that moles of gas present in the cylinder is 0.34 moles.

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a. The limiting reactant is NaHCO_{3}

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}⇒3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}

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By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of NaHCO_{3} react with 1 mole of H_{3} C_{6} HO_{7}  Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

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You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:  

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So <u><em>the limiting reagent is sodium bicarbonate</em></u>.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of NaHCO_{3} react with 3 mole of CO_{2}. Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

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You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}

<u><em>grams of carbon dioxide=  0.73 g</em></u>

<u><em>Then, 0.73 g of carbon dioxide are formed.</em></u>

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

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But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

<em><u>So the grams of excess reactant that do not participate in the reaction are 0333 g.</u></em>

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