Question

1.) how many mole of oxygen atoms are present in 5.36g of CaCo3?

2.) how many moles of Cl atoms are there in 1.7g of CCl4

3). What is the mass of 2.4x10^29 particles of CaCo3?

4.) In the reaction below, (CH4+2O2----> CO2 +2H2O) how many moles of CH4 are required to react with 0.5 mol of oxygen gas to form products?

5). Consider the reaction of 3 mol of P4 and 10 mol of oxygen gas to produce P2O5.

a. write the balance equation

b. how many moles of P2O5 could be produced from the reaction of 3 mol of P4 and 10 mol O2?

c.how many moles of excess reactant remain after the reaction is done?

Answer 1

1. Moles of CaCO3 = mass / molecular mass = 5.36 / 100 = 0.0536 moles

one mole of CaCO3 contain 3 mols of oxygen atoms

therefore 0.0536 moles of CaCO3 will contain = 0.0536 X 3 = 0.1608 moles

2. Moles of CCl4 = 1.7 / 153.8 = 0.011 Moles

one mole of CCl4 contain 4 moles of Cl atoms

0.011 mols of CCl4 will contain = 4 X 0.011 = 0.044 moles of Cl atoms

3. Mass of one mole or 6.022 X 10²³ particles of CaCO3 = 100

mass of one particle = 100 / 6.022 X 10²³ = 16.60 X 10^-23g

mass of 2.4x10²⁹ particles = 16.60 X 10^-23 X 2.4x10²⁹ = 33.2115 X 10⁶ g

4. CH4+2O2----> CO2 +2H2O

from above equation it is clear that one mole of CH4 reacts with 2 moles of Oxygen

2 moles of oxygen = 1 mole of CH4

1 mole of oxygen = 1 / 2 = 0.5 moles of CH4

therefore 0.5 moles of oxygen will react with 0.5 X 05 = 0.25 moles of CH4

5 a0 Balanced equation

P4 + 5O2 ----. 2P2O5

b) From above balanced reaction One mole of P4 reacts with 5 moles of O2 to form 2 moles of P2O5.

Therefore 3 moles of P4 will require 15 moles of O2 to react, but there are only 10 moles of O2, hence all the P4 will not react. Two moles of P4 will react with 10 moles of O2 hence one mole of P4 will remain unreacted.

Further one mole of P4 forms 2 moles of P2O10. If we double the moles of P4 from 1 to 2 the moles of product will also double from 2 to 4

therefore 4 moles of P2O10 will be formed.