Respuesta :
Answer:
56.0 grams NaCl
Explanation:
First, determine the limiting reactant by converting either one of the reactants' available quantities into the other reactant using the balanced equation. I picked 33.0 g Na:
33.0 g Na x 1 mol Na/22.99 g Na = 1.435 mol Na
1.435 mol Na x 1 mol Cl2/2 mol Na = 0.718 mol Cl2
0.718 mol Cl2 x 70.906 g Cl2/1 mol Cl2 = 50.89 g Cl2
As we can see, we only have 34.0 g of Cl2 to react with, though 33.0 g of Na can react with 50.89 g of Cl2. This means that Cl2 is our limiting reactant.
Now we can work off of the limiting reactant and stoichiometrically determine the grams of NaCl produced:
34.0 g Cl2 x 1 mol Cl2/70.906 g Cl2 = 0.4795 mol Cl2
0.4795 mol Cl2 x 2 mol NaCl/1 mol Cl2 = 0.9590 mol NaCl
0.9590 mol NaCl x 58.44 g NaCl/1 mol NaCl = 56.04 g NaCl
Round 56.04 g NaCl to three significant figures -> 56.0 g NaCl
The amount 56 gram of NaCl will be produced from 33.0 g of Na and 34.0 g of Cl.
What is gram?
The gram is defined as the mass of one cubic centimeter of pure water at 4 degrees Celsius
Calculation of amount of NaCl.
The given reaction is 2Na + Cl2 → 2NaCI.
Mass of Na = 33 gram
Mass of Cl = 34.0 gram
Conversion of gram into mol.
33 gram Na × 1 mole Cl2 / 22.99 g Na = 1.435 mol Na.
1.435 mol Cl2 × 70.906 gram Cl2 mol Na = 0.718 mol Cl2
0.718 mol Cl2 × 70.906 gram Cl 2 = 50.89 Cl2.
It can be seen that here, Cl2 is a limiting reagent.
Calculation of NaCl in gram
34 gram of Cl2 × 1 mol Cl2/ 70.906 g Cl2 = 0.4795 mol Cl2.
0.4795 mol Cl2 × 2 mol NaCl / 1 mol Cl2 = 0.9590 mol NaCl
0.9590 mol NaCl × 58.44 gram NaCl / 1 mol NaCl = 56 gram NaCl.
Therefore, the 56 gram of NaCl will be produced from 33.0 gram of Na and 34.0 gram of Cl.
To know more about gram.
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