Question

URGENT** If you have 35.0g of Na and 100.0g of Cl2, what is the maximum amount of NaCl that you can produce?

Answer 1

Answer:

We can produce 89.0 grams of NaCl

Explanation:

Step 1: Data given

Mass of Na = 35.0 grams

Mass of Cl2 = 100.0 grams

Molar mass of Na = 22.99 g/mol

Molar mass Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Na + Cl2 → 2NaCl

Step 3: Calculate moles Na

Moles Na = mass Na / molar mass Na

Moles Na = 35.0 grams / 22.99 g/mol

Moles Na = 1.52 moles

Step 4: Calculate moles Cl2

Moles Cl2 = mass Cl2 / molar mass Cl2

Moles Cl2 = 100.0 grams / 70.9 g/mol

Moles Cl2 = 1.41 moles

Step 5: Calculate limiting reactant

For 2 moles Na we need 1 mol Cl2 to produce 2 moles NaCl

Na is the limiting reactant. It will completely be consumed ( 1.52 moles).

Cl2 is in excess. There will react 1.52/2 = 0.76 moles

There will remain 1.41 - 0.76 = 0.65 moles

Step 6: Calculate moles of NaCl

For 2 moles Na we need 1 mol Cl2 to produce 2 moles NaCl

For 1.52 moles Na we'll have 1.52 moles NaCl

Step 7: Calculate mass NaCl

Mass NaCl = 1.52 moles* 58.44 g/mol

Mass NaCl =  88.8 grams ≈ 89 grams

We can produce 89.0 grams of NaCl

Answer 2

Answer:

165 g of NaCl are formed in the reaction

Explanation:

2Na + Cl₂ → NaCl

In order to determine the limiting reactant, we convert the mass of each reactant to moles

35 g / 23g/mol = 1.52 moles Na

100 g / 70.9 g/mol = 1.41 moles Cl₂

1 mol of chlorine reacts with 2 moles of Na, so If I have an x value of moles of Cl₂ I would need the double to react.

For 1.41 moles of Cl₂, I need 2.82 moles of Na; therefore my limiting reagent is the Na. Ratio is 2:2. So if I have 2.82 moles of Na I will produce 2.82 moles of NaCl

We convert the moles to mass: 2.82 mol . 58.45 g/1 mol =164.8 g