The fully hydrated form of sodium sulfate is the decahydrate, na2so4 • 10h2o. when heated the hydrated salt loses water. how many water molecules are found per formula unit in a partially dehydrated sample of sodium sulfate with a formula mass of 197 amu (i. e. find n for na2so4 • nh2o)?

There are found 3 water molecules per formula unit in a partially dehydrated sample of sodium sulfate with a formula mass of 197 amu.

Explanation:

The fully hydrated form of sodium sulfate is the decahydrate has structural formula of Na2SO4*10H2O

The molar mass of the decahydrate is 323 amu

When heated the hydrated salt loses water.

The molecular mass is now 197 amu

Na2SO4 has a molecular mass of 142.04 amu

1 molecule of water has a molecular mass of 18 amu

Before heating, the mass of Na2SO4*10H2O = 323 amu

After heating, the mass of Na2SO4*nH20 = 197 amu

The mass of nH2O  lost = 323-197 = 126 amu

1* H2O has a molecular mass of 18 amu

n = 126/18 = 7

The mass after heating - the mass of Na2SO4 = The mass of nH2O

197 - 142 = 55

The mass of nH2O / the mass of 1*H2O = n

55/ 18 = 3

This partially dehydrated sample of sodium sulfate is Na2SO4*3H2O

There are found 3 water molecules per formula unit in a partially dehydrated sample of sodium sulfate with a formula mass of 197 amu.

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