Question

You form water vapor by mixing oxygen and hydrogen at 730°C in a 5.4-liter container. This is the equation for the reaction: O2(g) + 2H2(g) → 2H2O(g). The partial pressure of oxygen before the reaction is 122.3 kilopascals, and there is excess hydrogen. How many moles of water are formed? The reaction produces moles of water.

Answer 1

The reaction is

$O_{2}(g)+2H_{2}(g)-->2H_{2}O(g)$

Thus from each mole of oxygen we will get two moles of water, if we are taking excess of hydrogen. As we are taking hydrogen in excess the oxygen is the limiting reagent

Let us calculate the moles of oxygen used from ideal gas equation

PV=nRT

where

P = pressure = 122.3 kPascals = 1.21 atm

V = volume = 5.4 L

T = temperature =$730^{0}C$=1003K

n=moles=?

R=gas constant = 0.0821 Latm / mol K

putting values

$moles=\frac{PV}{RT}=\frac{1.21X5.4}{0.0821X1003}=0.079mol$

therefore moles of water formed will be twice of moles of oxygen taken

$mole_{water}=2X0.079=0.158mol$

Answer 2

Answer:

0.158

Explanation: