Solid sodium reacts violently with water, producing Heath hydrogen gas, and sodium hydroxide ( 2Na + 2H2O --> 2NaOH + 2H2 ). How many molecules of hydrogen gas are formed when 67.2 g of sodium are added to water?
Here is a worked example of a stoichiometry problem. When you obtain moles H2 gas, convert to molecules by remembering that 1 mole of H2 gas contains 6.02E23 molecules.
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To determine the number of molecules of hydrogen gas formed when 67.2 g of sodium reacts with water, we need to use the given balanced chemical equation:
2Na + 2H2O -> 2NaOH + 2H2
The molar mass of sodium is 22.99 g/mol. We can calculate the number of moles of sodium using the following formula:
moles of sodium = mass of sodium / molar mass
moles of sodium = 67.2 g / 22.99 g/mol
moles of sodium = 2.92 mol
According to the balanced equation, the stoichiometric ratio between sodium and hydrogen gas is 2:2. This means that for every 2 moles of sodium, 2 moles of hydrogen gas are formed.
Since we have 2.92 moles of sodium, we can expect to produce the same number of moles of hydrogen gas. Therefore, the number of moles of hydrogen gas is also 2.92 mol.
Finally, we need to convert moles of hydrogen gas to molecules. One mole of any substance contains 6.022 x 10^23 molecules (Avogadro's number).
number of molecules of hydrogen gas = moles of hydrogen gas * Avogadro's number
number of molecules of hydrogen gas = 2.92 mol * 6.022 x 10^23 molecules/mol
So, the number of molecules of hydrogen gas formed when 67.2 g of sodium reacts with water is approximately 1.757 x 10^24 molecules.