Question

Sodium and water react according to the reaction 2Na(s) + 2H20(l) →2NaOH(aq) + H2(g) What number of moles ofH2 will be produced when 4 mol Na is added to 2 mol H20?​

Answer 1

Answer:

1 mol of Hydrogen (H2) produced when 4 mol Na is added to 2 mol H2O

Answer 2

Answer:

1 mole of hydrogen is produced.

Explanation:

$2Na(s) + 2H_{2}O(l)$ → $2NaOH(aq) + H_{2}(g)$

⇒2 moles of sodium atoms react with 2 moles of water molecules to give 2 moles of sodium hydroxide and 1 mole of hydrogen molecules.

⇒First, we need to find out the limiting reagent and excess reactant by dividing the given moles of each reactant by its coefficient.

⇒In the question, we have 2 moles of sodium atoms reacting with 2 moles of water molecules to produce a certain moles of hydrogen molecules.

$Na- \frac{4}{2} = 2$

$H_{2}O- \frac{2}{2} = 1$

⇒We can see that $Na$ is the excess reactant while $H_{2}O$ is the limiting reagent.

⇒Now we can calculate the amount of $H_{2}$ produced.

$H_{2}O : H_{2} \\2 : 1\\2 : x$

$2 * 1 = 2 * x$

$2 = 2x$

$1 = x$

Conclusion:

1 mole of hydrogen will be produced when we add 2 moles of water to 4 moles of sodium.