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GarryVolchara [31]

3 years ago

12

If 15 grams of Carbon dioxide is produced in a chemical reaction, how many grams of Carbon must be consumed in the reaction if w

e know there were 11 grams of Oxygen on the reactants side of the equation? (answer with only a number)

1 answer:

Nata [24]3 years ago

4 0

4.1g

Explanation:

Given parameters:

Mass of carbon dioxide = 15g

Mass of oxygen gas = 11g

Unknown:

Mass of carbon consumed = ?

Solution:

Equation of the reaction:

C + O₂ → CO₂

To solve this problem from the balanced equation, we have to use the amount of product formed and work to Carbon. This is because, we are sure of the amount of carbon dioxide formed but the amount of the given oxygen gas used is not precise.

Number of moles of CO₂ = $\frac{mass}{molar mass}$

Molar mass of CO₂ = 12 + (16 x2) = 44g/mol

Number of moles of CO₂ = $\frac{15}{44}$ = 0.34mole

From the equation of the reaction;

1 mole of CO₂ is produced from 1 mole of C

0.34mole of CO₂ will produce 0.34mole of C

Mass of carbon reacting = number of moles x molar mass = 0.34 x 12 = 4.1g

Learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

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in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

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Answer:

(a) False;

(b) False;

(c) False;

(d) True.

Explanation:

(a) When equilibrium is reached, the forward reaction rate becomes equal to the reverse reaction rate, that's why the molarity of each species remains constant, but reactions don't stop.

(b) According to the principle of Le Chatelier, an increase in molarity of either reactants or products would lead to a disturbance of equilibrium. This disturbance would lead to the shift of equilibrium towards the side which would minimize such a disturbance.

(c) Equilibrium constant is only temperature-dependent, it's independent of molarity, pressure, volume etc. of any species present in the reaction.

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Answer:

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