Consider the following balanced reaction between hydrogen and nitrogen to form ammonia: 3h2(g) + n2(g)→2nh3(g)
how many moles of nh3 can be produced from 24.0 mol of h2 and excess n2?

the balanced equation for the formation of ammonia is as follows

3H₂ + N₂ --> 2NH₃

we are told that N₂ is in excess. this means that H₂ is the limiting reactant

limiting reactant is fully consumed in the reaction , amount of product formed depends on amount of limiting reactant present

stoichiometry of H₂ to NH₃ is 3:2

when 3 mol of H₂ reacts - 2 mol of NH₃ are formed

therefore when 24.0 mol of H₂ reacts - 2/3 x 24.0 = 16 mol of NH₃ are formed

16 mol of NH₃ are formed

explanation: no but i do not usually comment

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.