1.25 calculate reacting masses using experimental data and chemical equations

The key to these is to know the basics, and then understand how ratios in balanced equations work. You should be able to do the sort of thing in the example below.

Remember that the balancing numbers in the equation give you the mole ratios.

e.g.      0.23 g of sodium reacts with excess water. What mass of sodium hydroxide is formed, and what volume of hydrogen gas? (Volume of 1 mole of gas at rtp is 24,000 cm³).

2Na + 2H2O → 2NaOH + H2

Step 1  work out the number of moles of the substance you can, in this case Na.     

Moles of Na = 0.23 g ÷ 23 = 0.010 mol

Step 2  use the balancing numbers in the equation to work out the number of moles of the substance you want to find. 

Mole ratio Na : NaOH = 2 : 2, so 1 : 1, which means the same number of moles of NaOH

Moles of NaOH = 0.010 mol

Step 3  convert the moles to whatever quantity the question asks.

Mass of NaOH = 0.020 ×  M_r   (23 + 16 + 1 =  40)

= 0.8 g

Now back to Steps 2 and 3 for the hydrogen part of the calculation.

Na : H2 = 2 : 1, so moles of hydrogen formed is HALF the number of moles of Na reacted

Moles of H2 = 0.010 ÷  2  =  0.005 mol

Volume of H2      = 0.005 × 24000 cm3

= 120 cm3

Such calculations should always be based on the amount of the reactant that is not in excess. This is sometimes called the yield-determining quantity.

sodium - soft enough to cut with a knife. Notice the metallic lustre of the freshly cut piece.