Phosphorus trifluoride is formed from its elements P4 (s) F2 (g) ---> PF3 (g) how many grams of fluorine are needed to react with 6.20 g

This is an incomplete question, here is a complete question.

Phosphorus trifluoride is formed from its elements.

How many grams of fluorine are needed to react with 6.20 g of phosphorus?

Answer : The mass of needed are, 11.4 grams.

Explanation : Given,

Mass of = 6.20 g

Molar mass of = 124 g/mol

Molar mass of = 38 g/mol

First we have to calculate the moles of

Now we have to calculate the moles of

The balanced chemical equation is:

From the balanced reaction we conclude that

As, 1 mole of react with 6 moles of

So, 0.05 moles of react with moles of

Now we have to calculate the mass of

Therefore, the mass of needed are, 11.4 grams.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g