Reacting masses and volumes
The masses and volumes of substances involved in a reaction can be determined by using equations.
In this case, the number of moles of a known substance is used to determine the amount of unknown substance by using the mole ratio in the equation. The equation must first be balanced with the correct formula for each substance. The number of moles of the known substance is always calculated first.
EXAMPLE 1
In an experiment, a solution containing 3.31g of lead (II) nitrate reacts with a solution containing 1.17g of sodium chloride to produce 2.78g of lead (II) chloride solid and leave a solution that contains 1.70g of sodium nitrate. What is the equation for the reaction? In this type of question, you are required to calculate the ratio of the reacting moles and then use these to write the equation. Pb(NO3)2 Mr = 331 NaCl Mr = 58.5 PbCl2 Mr =278 NaNO3 Mr = 85 3.31g of Pb(NO3)2 = (3.31/331)mol = 0.010mol 1.17g of NaCl = (1.17/58.5)mol = 0.020mol 2.78g of PbCl2 = (2.78/278)mol = 0.010mol 1.70g of NaNO3 = (1.70/85)mol = 0.020mol 0.01mol of Pb(NO3)2 reacts with 0.02mol of NaCl to give 0.01mol of PbCl2 and0.02mol of NaNO3 ie 1mol of Pb(NO3)2 reacts with 2mol of NaCl to give 1mol of PbCl2 and 2mol of NaNO3 Pb(NO3)2 + 2NaCl –– PbCl2 + 2NaNO3
EXAMPLE 2
When 5.175g of lead are heated at 300oC, the lead reacts with the oxygen in the air to produce 5.708g of an oxide of lead. This is the only product. What is the equation for this reaction?
In a question of this type, you seem to be short of information but, in fact, you know the mass of oxygen reacting. Remember, this is oxygen molecules that are reacting, not oxygen atoms. Mass of oxygen used is 5.708 - 5.175g = 0.533g Moles of lead reacting = (5.175/207) mol = 0.025mol Moles of oxygen reacting = (0.533/32) mol = 0.0167mol 0.025mol of Pb react with 0.0167mol of O2 to give product 1.5mol of Pb react with 1mol of O2 to give product 3mol of Pb react with 2mol of O2 to give product 3Pb + 2O2 –– Pb3O4
EXAMPLE 3
a. What mass of magnesium oxide would be produced from 16g of magnesium in the reaction between magnesium and oxygen? i Write the full-balanced equation. 2Mg(s) + O2(g) –– 2MgO(s) ii Read the equation in terms of moles. 2 moles of magnesium react to give 2 moles of magnesium oxide iii Convert the moles to masses using the Mr values (2 x 24g) of magnesium gives 2 x (24+16) = 80g of magnesium oxide 48g of magnesium gives 80g of magnesium oxide 16g of magnesium gives Xg = 80* 16/48 = 26.7g of magnesium oxide b. What volume of oxygen would react with 16g of magnesium in the above reaction?
In this case the oxygen is a gas, so the volume of each mole is 24,000 cm3 (24 dm3) at room temperature and pressure and you do not have to worry about the molecular mass of the gas. Remember: 1 dm3 = 1,000 cm3 From the equation: 2 moles of Mg reacts with 1 mole of O2 2 x 24g of Mg reacts with 1 x 24,000 cm3 of O2(g) 48g of magnesium reacts with 24, 000 cm3 of oxygen 16g of Mg reacts with X cm3 = 24,000*16/48 = 8,000 cm3 of oxygen
EXAMPLE 4
What mass of lead(II) sulphate would be produced by the action of excess dilute sulphuric acid on 10g of lead nitrate dissolved in water? Pb(NO3)2(aq) + H2SO4(aq) –– PbSO4(s) + 2HNO3(aq) 1 mole of lead nitrate gives 1 mole of lead sulphate 331g of lead nitrate gives 303g of lead sulphate 10g of lead nitrate gives Xg lead sulphate = 303 * 10/331g = 9.15g of lead sulphate