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Question

Question: A mixture of flammable gas containing carbon oxide (carbon (II) oxide) and hydrogen is produced from the reaction of water vapour with hot carbon. Calculate the total volume of carbon oxide and hydrogen produced by the reaction of 1.0t of carbon with excess water vapour at 10 degrees Celsius and 500kPa

I keep working this out and end up with a number that couldn't possibly be right.. Can someone solve this for me, showing their work, so I can use that to understand how to solve the other questions?

Answer 1

Carbon(II) oxide = CO

Hydrogen = H2

H20(g) + C(s) ==> CO(g) + H2(g) ... balanced equation

Assuming that 1.0t of carbon means 1.0 tons of carbon, we can proceed as follows:

moles of C(s) present = 1.0 tons x 2000 lbs/ton x 454 g/lb x 1 mol C/12 g = 75,667 moles C

moles CO produced = 75,667 moles C x 1 mol CO/mol C = 75,767 moles CO(g)

moles H2 produced = 75,667 moles C x 1 mol H2/mole C = 75,667 moles H2(g)

TOTAL MOLES OF GAS = 2 x 75,667 = 151,334 moles of gas formed

To find the total volume occupied by the gases, we use the ideal gas law PV = nRT and solve for V:

V = nRT/P

V = volume = ?

n = moles = 151,334

R = gas constant = 8.314 L-kPa/Kmole (note units are consistent with the given pressure)

T = temperature in K = 10ºC + 273 = 283K

P = pressure = 500 kPa

V = (151334)(8.314)(283) / 500

V = 712,136 L = 710,000 L (to 2 significant figures based on 1.0 ton)