Question

Find mL of gaseous NO when P=730 torr, T=298 K, with 100% yield.

Formlula: 3Pt+4HNO3+18HCl -> 3H2PtCl6+8H2O+4NO

Answer 1

Is there any other information? such as reactant concentrations or something else.

The problem alludes to using the ideal gas law but you don't have enough variables.

Answer 2

What other information is needed in order to solve this problem?

Answer 3

so if you were to use the ideal gas law, ultimately, you would need to know the mass of NO. This could be given indirectly, by giving you the initial mass (or concentration) of any of the reactants or the final mass (or concentration) of any of the other products - and you might need to do some stoichiometry.

Answer 4

That's also assuming that the pressure is solely attributed to NO.

Answer 5

It is true, the earlier questions has given 10 g of Platinum, I assume I need to find the limiting reactant, then find the mass for NO after. What then?

Answer 6

so when you find the mass of NO, you convert to moles and use the ideal gas law,

PV=nRT

and solve for V, volume.

Answer 7

They have given T and P, for n and R I'm not really sure how to get those, R specifically has multiple values and I don't know which one to use.

Answer 8

so the T given in already in Kelvin, the P is in torr, from here

https://en.wikipedia.org/wiki/Gas_constant

we can find the right units, R = 62.363 L torr/K mol

Normally i convert the pressure by dividing the P in torr by 760, which gives units in atmospheres.

n is moles, \(\sf moles=\dfrac{mass}{Molar~mass}\)

Answer 9

So it would be something like \[V=\frac{ (NO)(62.363)(298) }{ atm}\]

Also when I find the limiting reactant, I can use that to find the NO directly? Like a product which is the limiting reaction, stoichiometry that with NO?

Answer 10

also NO in the equation is mols of NO

Answer 11

The equation is mostly correct, the value of 62.363 was when you're using the pressure in units of torr. The atm comment i made is what i normally do to avoid memorizing all the values for R. So the equation should look like this:

\(\sf V=\dfrac{nRT}{P}=\dfrac{n*(62.363 ~L ~torr/K~ mol)*(298~K)}{ 730~torr}\)

The limiting reactant will give you the highest possible amount of moles for the reaction, so yes by using the moles of the limiting reactant (and the stoichiometric coefficients of the limiting reactant and that of NO) you can find the moles of NO.

Answer 12

You're a god. Thank you for everything. I really appreciate it. I think I can handle it from here. Thanks again.

Answer 13

lol hardly, but thanks.

No problem, glad i could clear things up!