If 22.5 liters of oxygen reacted with excess of hydrogen, how many liters of water vapor could be produced?

45 L

Explanation:

Assuming normal temperature and pressure (NTP) and ideal gas behaviour, 22.5 L are equivalent to:

PV = nRT

n = PV/RT

n = (1*22.5)/(0.082*273.15)

n = 1 mol

Oxygen and hydrogen react to form water as  follows:

2 H2 (g) + O2 (g) -> 2 H2O (g)

From the balanced equation, we know that 1 mol of Oxygen produces 2 moles of water. The volume occupied by these 2 moles is:

PV = nRT

V = nRT/P

V =  2*0.082*273.15/1

V = 45 L

Know we can see the assumption of NTP conditions was unnecessary, the ratio between moles of oxygen consumed and moles of water produced is 1:2, so the ratio between volumes.

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.