Your answer of 150 L is correct but I don't agree with anything else you said. What do you mean you have no way to obtain n = mols.
PV = nRT or n = PV/RT. For H2 you know P, V, R and T so you can calculate n for H2, convert to mols O2, then use pV = nRT and solve for V in L. You get the same 150 L but you went around the barn to get it. That's why someone thought of the law of combining gases. All of that "other stuff" cancels so that using the volume works the same (BUT MUCH QUICKER) as using mols. You leave out all that R, T, P stuff AS LONG AS YOU KNOW THE BEGINNING AND ENDING p AND T ARE THE SAME
what volume of gas at 40°C and 1.50 atm is necessary to react with 300 L of hydrogen gas measured at the same conditions?
2H2 (g) + O2 (g) --> 2H20 (g)
1.50 atm= 151.KPa
How would I solve this? Because I can't use: PV = nRT
I have pressure, temperature, r. There's no way I can find moles and volume. I need the moles.
So would I just do:
300L H2× 1 L O2 / 2 L H2
And I get 150L as the volume for Oxygen by using law of combining gases.
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