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Determination of the relative atomic mass of lithium Results Results of first procedure: 158cm3 of hydrogen gas collected Results of second procedure Try Amount of acid needed (cm3) 1 39.20 2 39.30 3 39.30 Average 39.27 Conclusion Balanced Equation = 2Li(s) + 2H2O(l) › 2LiOH(aq) + H2(g) *All results will be rounded up to three significant figures Method 1 Assume that 1 mole of gas occupies 24000cm3 at room temperature and pressure. Firstly, the number of moles of Hydrogen gas needs to be calculated. This is done by dividing the volume of hydrogen gas produced in the experiment (This would be 158cm3) by the volume of 1 mole of hydrogen gas (24000cm3): Number of moles of H2 gas = Volume of H2 (g)/Volume of 1 mole of H2 (g) 158cm3/24000cm3 = 0.0065833 The result of this calculation is = 0.00658 The mole ratio of Lithium to Hydrogen is worked out by the balanced equation of the experiment, as shown above. For every 1 mole of hydrogen, there is going to be 2 moles...

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