A chemist carries out a reaction between 10.0g of sodium and 20.0g chlorine. Which reactant element is present in excess? What quantity of sodium chloride is produced? What quantity of the elment in short supply would have been required to react with all of the element present in exress?

write a balance equation.

2Na+ Cl2>>2NaCl

now, calculate mole you have of each:
Cl2moles=20/70=.288
Namoles=10/23 =.435
you need from the equation, twice the moles of Na as Cl2
so you are short Na.
You take it from here.

To determine which reactant element is present in excess, we need to compare the stoichiometry of the reaction to the amounts of each reactant given.

Step 1: Write the balanced equation for the reaction between sodium and chlorine:
2Na + Cl2 → 2NaCl

Step 2: Calculate the number of moles for each reactant:
Moles of sodium (Na) = mass / molar mass = 10.0g / 22.99g/mol ≈ 0.435 mol
Moles of chlorine (Cl2) = mass / molar mass = 20.0g / 35.45g/mol ≈ 0.563 mol

Step 3: Determine the mole ratio of the reactants according to the balanced equation:
2 moles Na : 1 mole Cl2

Step 4: Compare the reactant mole ratios:
The mole ratio of Na to Cl2 is 2:1, which means that for every 2 moles of Na, we need 1 mole of Cl2.

Since we have more moles of Cl2 (0.563 mol) than the ratio requires (0.435 mol), chlorine (Cl2) is the element present in excess.

Now, let's determine the quantity of sodium chloride (NaCl) produced:

Step 5: Determine the mole ratio of the product (NaCl) to the limiting reactant:
From the balanced equation, we can see that 2 moles of Na react to form 2 moles of NaCl.

Step 6: Calculate the moles of NaCl produced:
Moles of NaCl = moles of limiting reactant (Na) = 0.435 mol

Step 7: Calculate the mass of NaCl produced:
Mass of NaCl = moles of NaCl * molar mass = 0.435 mol * (22.99g/mol + 35.45g/mol) ≈ 23.0g

So, approximately 23.0 grams of sodium chloride are produced.

Finally, let's determine the quantity of the element in short supply (Na) required to react with all of the element present in excess (Cl2):

Step 8: Determine the mole ratio of the element in short supply (Na) to the excess element (Cl2):
From the balanced equation, we can see that 2 moles of Na react with 1 mole of Cl2.

Step 9: Calculate the moles of Na that react with the excess Cl2:
Moles of Na = (moles of Cl2) * (2 moles Na / 1 mole Cl2) = 0.563 mol * (2/1) = 1.126 mol

Step 10: Calculate the mass of Na required to react with the excess Cl2:
Mass of Na = moles of Na * molar mass = 1.126 mol * 22.99g/mol ≈ 25.9g

Therefore, approximately 25.9 grams of sodium would have been required to react with all of the chlorine present in excess.