Question

A student carefully placed 15.6 g of sodium in a reactor supplied with an excess quantity of chlorine gas. When the reaction was complete, the student obtained 39.7 g of sodium chloride. Calculate how many grams of chlorine gas reacted. How many grams of sodium reacted?

Answer 1

Answer:

15.6 g of sodium and 24.1 g of chlorine react to produce 39.7 g of sodium chloride.

Explanation:

Given data:

Mass of sodium = 15.6 g

Mass of sodium chloride = 39.7 g

Mass of chlorine react = ?

Mass of sodium react = ?

Solution:

Balance chemical equation:

2Na + Cl₂ → 2NaCl

Now we will calculate the moles of sodium reacted.

Number of moles = mass /molar mass

Number of moles =  15.6 g / 23 g/mol

Number of moles =  0.68 mol

Now we will calculate the moles of sodium chloride produced,

Number of moles of NaCl = 39.7 / 58.5 g/mol

Number of moles of NaCl = 0.68 mol

Since 0.68 moles of sodium produced 0.68 moles of sodium chloride or we can say 2 moles of sodium produced 2 moles of sodium chloride.

Now we calculate the mass of chlorine reacted,

From balance chemical equation we compare the moles of chlorine and sodium

           Na      :    Cl

             2       :     1

            0.68   :     1/2× 0.68 = 0.34 mol

Mass of chlorine reacted = moles × molar mass

Mass of chlorine reacted = 0.34 mol × 71 g/mol = 24.1 g

According to the law of conservation of mass,

2Na + Cl₂ → 2NaCl

15.6 g + 24.1 g = 39.7 g

15.6 g of sodium and 24.1 g of chlorine react to produce 39.7 g of sodium chloride.

Answer 2

Answer:

15.6g of sodium reacted, 24.1 g of chlorine gas reacted.

Explanation:

Law of Conservation | Mass of reactants = Mass of Products

15.6 + c = 39.7

-                 -

c = 24.1