Answer:
Number of moles of hydrogen gas produced when 15.3 g of sodium reacts with water = 0.333 moles of hydrogen gas
Explanation:
The reaction between sodium metal and water is given by the chemical equation below:
2Na(s) + 2H₂O(l) ------> 2NaOH(aq) + H₂(g)
From the equation of reaction above, 2 moles of sodium reacts with 1 mole of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen gas.
mole ratio of sodium and hydrogen gas is 2:1
molar mass of sodium =23 g/mol:
number of moles of sodium present in 15.3 g = mass/molar mass
number of moles of sodium present in 15.3 g = 15.3 g/ 23 g/mol = 0.665 moles
number of moles of hydrogen gas produced = 0.665 * 1/2 = 0.333 moles
Therefore, number of moles of hydrogen gas produced when 15.3 g of sodium reacts with water is 0.333 moles
