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C 3 H 6 (g) + O 2(g) --------------> CO 2(g) + H 2 O (g)
What volume of C 3 H 6 to produce 34.7 L of H 2 O at a temperature of 29.4◦ C and a pressure of 0.762 atm?

Using the equation above solve the following: How many liters of CO 2 can be produced from 18.7 g of O 2 at STP?

Respuesta :

As per ideal gas equation ,

 [tex]C_{3}H_{6}[/tex] Volume   = 11.56 L

[tex]CO_{2\\}[/tex] volume =  8.62L

What is Ideal gas equation?

PV = nRT

[tex]C_{3} H_{6}[/tex] gas burns in air according to following equation:

[tex]2C_{3}H_{6}[/tex] + [tex]9O_{2}[/tex] → [tex]6CO_{2}[/tex] + [tex]6H_{2} O[/tex]

Volume of [tex]C_{3} H_{6}[/tex] produced;

Step 1 : Calculate the moles of [tex]H_{2} O[/tex]

PV = nRT

n = PV / RT

n = 0.762 × 34.7 / 0.08206 × 302.55

n = 1.065 mol

Step 2: Calculate the moles of [tex]C_{3} H_{6}[/tex]

1.065 mol [tex]H_{2} O[/tex]× 2 mol [tex]C_{3} H_{6\\}[/tex] / 6 mol [tex]H_{2} O[/tex]

= 0.355 mol

Step 3 : Calculate new volume

V = 0.355 × 0.08206 × 302.55 / 0.762

V = 11.56 L

Volume of [tex]CO_{2}[/tex] produced;

Step 1: Calculate the moles of [tex]O_{2}[/tex]

18.7g [tex]O_{2}[/tex] × 1 mol [tex]O_{2}[/tex] / 32 g [tex]O_{2}[/tex]

= 0.58 mol

Step 2: calculate the moles of [tex]CO_{2}[/tex]

0.58 mol × 6mol [tex]CO_{2}[/tex] / 9 mol [tex]O_{2}[/tex]

= 0.38 mol

Step3: convert mol to litres

0.38 × 22.7L / 1mol [tex]CO_{2}[/tex]          [22.7L = 1mol]

= 8.62 L

Learn more about ideal gas here:

https://brainly.com/question/4147359

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