All categories

3 years ago

In the reaction 2CO (g) + O2 (g) 2CO2 (g), how many moles of oxygen gas are needed to react with 24 moles of carbon monoxide

Chemistry

1 answer:

Anuta_ua [19.1K]3 years ago

4 0

12 moles of oxygen gas are needed to react with 24 moles of carbon monoxide.

Explanation:

The molar ratio of carbon monoxide to oxygen 2:1

Which means 2 moles of carbon monoxide is reacting with 1 mole of oxygen.

to produce 2 moles of carbon dioxide.

Therefore, from the molar ratio, we get that 12 moles of oxygen are required to react with 24 moles 0f carbon monoxide.

Molar ratio:

The molar ratio gives the moles of product that are formed from a certain amount of reactant, and also the number of moles of a reactant needed to react with another reactant.

You might be interested in

A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0

3 years ago

A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24

Setler79 [48]

Answer:

For Part A: The partial pressure of Helium is 218 mmHg.

For Part B: The mass of helium gas is 0.504 g.

Explanation:

For Part A:

We are given:

$p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg$

To calculate the partial pressure of helium, we use the formula:

$P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}$

Putting values in above equation, we get:

$745=245+119+163+p_{He}\\p_{He}=218mmHg$

Hence, the partial pressure of Helium is 218 mmHg.

For Part B:

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

$218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g$

Hence, the mass of helium gas is 0.504 g.

6 0

3 years ago

What is the Celsius temperature of 1 mole of a gas that has an average kinetic energy of 4,290 joules?

timofeeve [1]

Well the the answer is 70.8c but if you round it up it is 71c which I choice and got it correct so the answer is 71c

5 0

3 years ago

Someone help me please please help whats the ratio of carbon and hydrogen???????

MatroZZZ [7]

1:2

The ratio of carbon, hydrogen, and oxygen in most carbohydrates is 1:2:1. This means for every one carbon atom there are two hydrogen atoms and one...

3 0

3 years ago

Calculate Ho298 for the process

Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}$ $\Delta H^0_1 = -314 kJ$ ..........(1)

$SbCl_{3} + Cl_2 \rightarrow SbCl_{5}$ $\Delta H^0_2 = -80kJ$ ..............(2)

The final reaction is as follows:

$Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}$ $\Delta H^0_3 = ?$ .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of $\Delta H^{0}_{3}$ at 298 K will be as follows.

$\Delta H^{0}_{3}$ = $\Delta H^{0}_{1}$ + $\Delta H^{0}_{2}$

= -314 kJ + (-80) kJ

= -394 kJ

Thus, we can conclude that $H^{o}$ at 298 K for the given process is -394 kJ.

4 0

3 years ago