if 6.506.50 l of water vapor at 50.2 °c50.2 °c and 0.1210.121 atm reacts with excess iron, how many grams of iron(iii) oxide will be produced? 2fe(s) 3h2o(g)⟶fe2o3(s) 3h2(g)

Respuesta :

V = 6.50 L

T = 50.2 degree celsius = 50.2 + 273.15 K = 323.35 K

P = 0.121 atm

Using ideal gas equation,

n = PV/ RT = 0.121 atm* 6.50 L / 0.0821 * 323.35 K = 0.0296 mol

0.0296 mol of H20* 1mol Fe2O3 / 3mol H2O * 159.69g of Fe2O3/1 mol Fe2O3= 1.57g

So, 1.57g of iron(iii) oxide will be produced.

What do you mean by ideal gas equation?

The Ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas equation can be written as

PV = nRT

Where,

P is the pressure of the ideal gas.

V is the volume of the ideal gas.

n is the amount of ideal gas measured in terms of moles.

R is the universal gas constant.

T is the temperature.

To know more about ideal gas equation from the given link:

https://brainly.com/question/20348074

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