How many grams of NaF form when .5 mol of HF reacts with excess Na2SiO3?

5.25g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is shown below:

Na2SiO3 + 8HF → H2SiF6 + 2NaF + 3H2O

From the balanced equation above,

8 moles of HF reacted to produce 2 moles of NaF.

Therefore, 0.5 moles of HF will react to produce = (0.5 x 2)/8 = 0.125 mole of NaF.

Next, we shall convert 0.125 mole of NaF to grams.

This is illustrated below:

Mole of NaF = 0.125 mole

Molar mass of NaF = 23 + 19 = 42g/mol

Mass of NaF =..?

Mass = mole x molar mass

Mass of NaF = 0.125 x 42

Mass of NaF = 5.25g

Therefore, 5.25g of NaF is produced from the reaction.

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.