A sample of sodium sulfite has a mass of 2.80 g.

a) How many Na^+ ions are present?
b) How many SO_3^2- ions are present?
c) What is the mass in grams of one formula unit of Na_2SO_3?

For a: The number of sodium ions in given amount of sodium sulfite are

For b: The number of sulfite ions in given amount of sodium sulfite are

For c: The mass of one formula unit of sodium ions is grams

Explanation:

The chemical formula of sodium sulfite is  . It is formed by the combination of 2 sodium ions and 1 sulfite ions

To calculate the number of moles, we use the equation:

Given mass of sodium sulfite = 2.80 g

Molar mass of sodium sulfite = 126 g/mol

Putting values in above equation, we get:

Moles of sodium ions in sodium sulfite = (2 × 0.022) moles

According to mole concept:

1 mole of a compound contains number of particles

So, 0.022 moles of sodium sulfite will contain = number of sodium ions

Hence, the number of sodium ions in given amount of sodium sulfite are

Moles of sulfite ions in sodium sulfite = (1 × 0.022) moles

According to mole concept:

1 mole of a compound contains number of particles

So, 0.022 moles of sodium sulfite will contain = number of sulfite ions

Hence, the number of sulfite ions in given amount of sodium sulfite are

Molar mass of sodium sulfite = 126 g/mol

According to mole concept:

number of formula units are present in 1 mole of a compound

Or, number of formula units of sodium sulfite have a mass of 126 grams

So, 1 formula unit of sodium sulfite will have a mass of =

Hence, the mass of one formula unit of sodium ions is 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

14,16 18 reacts with group