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Suppose 275 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?

I attached the image of the equation for reference

Suppose 275 g of NO3 flows into a swamp each day What volume of CO2 would be produced each day at 170C and 100 atm I attached the image of the equation for refe class=

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Eduard22sly

Answer:

264.28 L of CO₂

Explanation:

We'll begin by calculating the number of moles present in 275 g of NO₃¯. This can be obtained as follow:

Mass of NO₃¯ = 275 g

Molar mass of NO₃¯ = 14 + (16×3) = 14 + 48 = 62 g/mol

Mole of NO₃¯ =?

Mole = mass / molar mass

Mole of NO₃¯ = 275 / 62

Mole of NO₃¯ = 4.44 moles

Next, we shall determine the number of mole of CO₂ produced from the reaction. This can be obtained as follow:

2NO₃¯ + 5CO + 2H⁺ —> N₂ + H₂O + 5CO₂

From the balanced equation above,

2 moles of NO₃¯ reacted to produced 5 moles of CO₂.

Therefore, 4.44 moles of NO₃¯ will react to produce = (4.44 × 5) / 2 = 11.1 moles of CO₂.

Thus, 11.1 moles of CO₂ were produced from the reaction.

Finally, we shall determine the volume of CO₂ produced each day. This can be obtained as follow:

Temperature (T) = 17 °C = 17 °C + 273 = 290 K

Pressure (P) = 1 atm

Number of mole (n) of CO₂ = 11.1 moles

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) of CO₂ =?

PV = nRT

1 × V = 11.1 × 0.0821 × 290

V = 264.28 L

Thus, 264.28 L of CO₂ would be produced each day.

The volume of [tex]\rm CO_2[/tex] produced has been 264.28 L.

The volume of the carbon dioxide gas produced can be calculated with the ideal gas equation.

The moles of [tex]\rm NO_3^-[/tex] has been:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Moles of 275 g  [tex]\rm NO_3^-[/tex] has been:

Moles = [tex]\rm \dfrac{275}{62}[/tex]

Moles of  [tex]\rm NO_3^-[/tex] = 4.44 moles

The balanced equation gives that with the reaction of 2 moles [tex]\rm NO_3^-[/tex] , 5 moles carbon dioxide has been formed.

Moles of carbon dioxide formed with 4.44 mol [tex]\rm NO_3^-[/tex] has been:

2 moles [tex]\rm NO_3^-[/tex] = 5 moles [tex]\rm CO_2[/tex]

4.44 moles [tex]\rm NO_3^-[/tex]  = 11.1 moles [tex]\rm CO_2[/tex].

The volume of the [tex]\rm CO_2[/tex] can be calculated as:

Pressure [tex]\times[/tex] Volume = moles [tex]\times[/tex] Rydberg constant [tex]\times[/tex] temperature

For the [tex]\rm CO_2[/tex] produced:

1 atm [tex]\times[/tex] Volume = 11.1 mol [tex]\times[/tex] 0.0821 atm.L/K.mol [tex]\times[/tex] 290 K

Volume = 264.28 L.

The volume of [tex]\rm CO_2[/tex] produced has been 264.28 L.

For more information about the volume of gas, refer to the link:

https://brainly.com/question/15265703

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