Question

Assuming all volume measurements are made at the same temperature and pressure how many liters of carbon dioxide gas can be produced 12.8 L of oxygen gas react with excess carbon monoxide

Answer 1

Answer:

25.6 liters of $CO_2$ will be produced.

Explanation:

We are given that 12.8 liters of Oxygen gas is reacted with excess Carbon Monoxide to produce of Carbon Dioxide gas, keeping all the volume measurements at the same temperature.

We are to find the number of liters of Carbon Dioxide used in the reaction.

The equation for this reaction will be:

$2 CO + O_2\implies 2 CO_2$

From the equation, we can see that the mole ratio of Oxygen to Carbon Dioxide is 1:2. Therefore, the number of liters of $CO_2$ will be double the number of liters of Oxygen.

$12.8 \times 2 = 25.6$

So, 25.6 liters of $CO_2$ will be produced.

Answer 2

25.6 L of carbon dioxide gas will be produced when 12.8 L of oxygen gas react with excess carbon monoxide. The stoichiometric ratio for oxygen gas and carbon dioxide gas is 1:2. For every mole of O2 consumed, 2 moles of CO2 are formed.

Further Explanation:

The balanced chemical equation for this reaction is:

O2 + 2 CO → 2 CO2

To get the amount of CO2 produced, the following steps must be followed:

1. Convert the volume of O2 to moles O2.
2. Get the number of moles of CO2 formed using the stoichiometric ratio 1 mole O2: 2 moles CO2
3. Convert the moles of CO2 produced to liters

Solving the problem,

STEP 1: Convert liters to moles

$moles \ of \ O_{2} \ = given \ volume \ O_{2} \ (\frac{1 \ mole \ O_{2}}{22.4 \ L} )\\moles \ of \ O_{2} \ = 12.8 \ L \ O_{2} \ (\frac{1 \ mole \ O_{2}}{22.4 \ L} )\\\boxed {moles \ of \ O_{2} \ = 0.57143 \ mol}$

STEP 2: Get the moles of CO2 formed

$moles \ of\ CO_{2} \ = \ given \ moles \ O_{2} \ (\frac{2 \ mol \ CO_{2}}{1 \ mol \ O_{2}})\\moles \ of\ CO_{2} \ = \ 0.57143 \ mol \ O_{2} \ (\frac{2 \ mol \ CO_{2}}{1 \ mol \ O_{2}})\\\\\boxed {moles \ of\ CO_{2} \ = \ 1.14286 \ mol}$

STEP 3: Convert moles to liters

$liters \ CO_{2} \ = given \ moles \ CO_{2} \ (\frac{22.4 \ L}{1 \ mol \ CO_{2}})\\liters \ CO_{2} \ = 1.14286 \ mol \ CO_{2} \ (\frac{22.4 \ L}{1 \ mol \ CO_{2}})\\\\\boxed {liters \ CO_{2} \ = 25.6 \ L}$

Learn More

1. Learn more about STP brainly.com/question/11676583
2. Learn more about molar volume brainly.com/question/4172228
3. Learn more about ideal gas law brainly.com/question/12968132

Keywords: stoichiometry, STP, gases