If 35.4 liters of hydrogen gas are used, how many liters of nitrogen gas will be needed for the above reaction at stp ​

11.8 L

Explanation:

Step 1: Write the balanced reaction between hydrogen and nitrogen

3 H₂(g) + N₂(g) = 2 NH₃(g)

Step 2: Calculate the moles corresponding to 35.4 L of H₂

At STP, 1 mole of H₂ occupies 22.4 L.

35.4 L × 1 mol/22.4 L = 1.58 mol

Step 3: Calculate the moles of N₂ required to react with 1.58 moles of H₂

The molar ratio of H₂ to N₂ is 3:1. The moles of N₂ required are 1/3 × 1.58 mol = 0.527 mol

Step 4: Calculate the volume occupied by 0.527 moles of N₂ at STP

At STP, 1 mole of N₂ occupies 22.4 L.

0.527 mol × 22.4 L/1 mol = 11.8 L

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.