Limiting Reactant?

Question

What mass of water in grams forms from the reaction of 2.04 L of hydrogen gas and 1.94 L of oxygen gas ? Both gases are at 655 torr and 20.0 degrees Celsius.

Answer 1

According to the chem. equation:

2H2 + O2 --> 2H2O

2.04 L of H2 require 1.02 L of O2, measured at the same pressure and temperature. We have 1.94 L of O2, more than we need, so H2 is the limiting reactant, reacts fully and produces 2.04 L of H2O(g), measured at p = 655 torr - or 655/760 atm and at 20 oC - or 293 K. Now you can find the moles of H2O:

pV = nRT => n = p*V/R*T = (655/760)*2.04/0.082*293 = 0.0732 mol

and then you can find the mass of H2O:

m = n*Mr = 0.0732*18 = 1.317 g or 1.32 g approx.

Answer 2

O2 + 2H2 --->2 H20

Balance that baby!

Use PV=nRT to find how many moles of each you have using the given liters of each, the 20 C converted to K, and the 655 torr converted to atm with a .08206 r value.

Now you have moles and you can calculate which is the limiting reactant.

Take the moles of O2 you calculate and multiply by 2 mole H2/1 mole O2 (molar ratio) and find the H2 REQUIRED. Do the same with the moles H2 present and multiply it by 1/2 to find the O2 REQUIRED....see how the units cancel out?? Whichever reactant has more required than present is the limiting reactant.

Take the number of moles of limiting reactant and multiply by the molar ratio of water divided by that element from the equation to find moles produced and simply use molar mass to get to grams

Answer 3

I want you to understand and not just try toget the anwer so I'll help a little.
Use pv=nRT to solve for n (moles) of each compound.
The mole ratio of hydrogen to oxgen in water is 2:1

Does that help?