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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?

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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!

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