Question

for the reaction represented by the equation ( 2na + 2h_2o \to 2naoh + h_2 ), how many grams of hydrogen are produced if 120.g of sodium and 80.g of water are available?
80.g
200 g
4.5 g
9
multiple choice 1 point
which reactant controls the amount of product formed in a chemical reaction?
limiting reactant
composition reactant
mole ratio

Explanation:

Step1: Calculate moles of reactants

Molar mass of \(Na = 22.99\ g/mol\), moles of \(Na=\frac{120.}{22.99}\approx5.22\ mol\)
Molar mass of \(H_2O = 18.02\ g/mol\), moles of \(H_2O=\frac{80.}{18.02}\approx4.44\ mol\)
From the equation \(2Na + 2H_2O
ightarrow2NaOH + H_2\), mole ratio of \(Na:H_2O = 1:1\). Since \(5.22>4.44\), \(H_2O\) is the limiting reactant.

Step2: Calculate moles of \(H_2\)

From the equation, mole ratio of \(H_2O:H_2 = 2:1\). Moles of \(H_2=\frac{4.44}{2}=2.22\ mol\)

Step3: Calculate mass of \(H_2\)

Molar mass of \(H_2 = 2.02\ g/mol\), mass of \(H_2=2.22\times2.02\approx4.5\ g\)

In a chemical reaction, the limiting reactant is the one that gets completely consumed first and determines the amount of product formed. Here, by calculating moles of \(Na\) and \(H_2O\) and comparing their mole - ratio with the reaction equation, \(H_2O\) is found to be the limiting reactant. Using the stoichiometry of the reaction (mole ratios), moles of \(H_2\) are calculated and then converted to mass.

Answer:

4.5 g