Need help with two chemistry questions dealing with stoichiometry

I'm stuck with these two questions.. can anyone help?

1. How many grams of Rubidium metal are needed to produce 12.7 x 10^5 liters of Hydrogen gas (h2 at STP)?

After balancing the equation, I have this reaction

2Rb(s) + 2HOH (l) ---> H2(g) + 2RbOH (aq)

I'm not sure what to do from here... can anyone show me?

The second question is

2. How many moles of Oxygen (O2) are required for the combustion of 45.7 moles of butene (C4H8)

I have this reaction that I have balanced

C4H8(s) + 6O2(g) ----> 4H2O(g) + 4CO2(g)

Can anyone show me how to do these two? Thank you.

1.
At STP 1 mole of any ideal gas occupies a volume of 22.4 L

So moles H2 at STP = volume (L) / 22.4 L/mol
= 12.7x10^5 L / 22.4 L
= 56696 mol

Now, your equation is fine (although generally we write water as H2O)
2Rb(s) + 2H2O(l) ----> 2RbOH(aq) + H2(g)

we can see that 2 mole Rb react to form 1 mole H2
So moles Rb required = 2 x moles H2 produced
= 2 x 56696 mol
= 113393 mol

mass = molar mass x moles
= 85.47 g/mol x 113393 mol
= 9691687.5 g
= 9.69x10^6 g (3 sig figs)


2. the balanced equation tells you that 6 moles O2 are required to combust 1 mole C4H8
So moles O2 needed = 6 x moles C4H8
= 6 x 45.7 mol
= 274.2 mol
= 274 mol (3 sig figs)

Arf