How many grams of chlorine are required to react completely with 5.00 grams of sodium to produce sodium chloride ? 2na + cl2 → 2nacl
2Na + Cl2 ==> 2NaCl
mols Na = grams/atomic mass
mols Cl2 = mols Na x 1/2 (Look at the coefficients in the balanced equation.)
grams Cl2 = mols Cl2 x molar mass Cl2.
Student High School
2Na(s) + Cl2(g)> 2NaCl(s)
Number of moles= mass/molar mass
Number of moles of sodium= 5/23
=0.213 moles
Mole ratio of sodium to chlorine is 2:1
therefore, number of moles of chlorine =1/2×0.213
=0.1065 moles
Mass= number of moles×molar mass
therefore, mass of chlorine= 0.1065×35.5
=3.78075g
To find out how many grams of chlorine are required to react completely with 5.00 grams of sodium, we need to determine which reactant is limiting in the given chemical equation.
First, let's calculate the molar masses of sodium (Na) and chlorine (Cl₂):
Molar mass of Na = 22.99 g/mol
Molar mass of Cl₂ = 2 * (35.45 g/mol) = 70.90 g/mol
Next, use the molar masses to convert grams to moles:
Moles of Na = 5.00 g / 22.99 g/mol ≈ 0.217 mol (rounded to three decimal places)
From the balanced equation, we can see the stoichiometric ratio between sodium and chlorine is 2:1. This means that for every 2 moles of sodium, we require 1 mole of chlorine.
Since the stoichiometric ratio is given in moles, we need to convert the moles of sodium to moles of chlorine using the ratio:
Moles of Cl₂ = (0.217 mol Na) / 2 mol NaCl₂ = 0.1085 mol Cl₂ (rounded to four decimal places)
Finally, convert the moles of chlorine to grams:
Grams of Cl₂ = Molar mass of Cl₂ * Moles of Cl₂
Grams of Cl₂ = 70.9 g/mol * 0.1085 mol Cl₂ ≈ 7.70 grams of Cl₂ (rounded to two decimal places)
Therefore, approximately 7.70 grams of chlorine are required to react completely with 5.00 grams of sodium to produce sodium chloride.