A 25.0-g sample of barium reacted completely with water. What is the equation for the reaction? How many milliliters of dry H2 evolved from 21C and 748mmHg?

Ba+H↓2O⇒BaO+H↓2O

You'll have to divide 25.0g by the formula mass of barium and then compare the mole ratio in your formula.

PV=nRT Rearranged is: V=nRT/P

V=volume=? n=moles of H2 P=pressure 748mmhg

R=universal gas constant (62.3638L*mmHg/K*mol)

T = temperature (21+273.15=) 294.15K

L=1000ml

25.0g divided by 137.3g (formula mass of barium) = moles for barium

1 mole of barium = 1 mole of H2

H2 moles times 62.3638 L*mmHg/K*mol (these cancel out to L) times 294.15K over 748mmHg

Divide your answer which should be in L to ml by multiplying it by 1000.

(25/137.3) (62.3638*294.15) / 748=answer*1000

I got 4.47x10∧3 but depending on how your teacher calls each number significant change your sig figs to the lowest number. Should look pretty similar. Good luck!