Question

Help please? This is my last question. Will give BRAINLIEST to correct answer. :)

Answer 1

Answer:

There will be 44.53 grams of sodium oxide produced

Explanation:

Step 1: Data given

Mass of sodium chloride (NaCl) = 84.00 grams

Magnesium oxide is in excess

Molar mass of NaCl = 58.44 g/mol

Molar mass of Na2O = 61.98 g/mol

Step 2: The balanced equation

2NaCl + MgO → Na20 + MgCl2

Step 3: Calculate moles of NaCl

Moles NaCl = Mass NaCl / molar mass NaCl

Moles NaCl = 84.00 grams / 58.44 g/mol

Moles NaCl = 1.437 moles

Step 4: Calculate moles of Na2O

The limiting reactant is NaCl.

For 2 moles of NaCl we need 1 mol of MgO to produce 1 mol of Na2O and 1 mol of MgCl2

For 1.437 moles of NaCl we'll have 1.437/2 = 0.7185 moles of Na2O

Step 5: Calculate mass of Na2O

Mass of Na2O = moles Na2O * molar mass Na2O

Mass Na2O = 0.7185 mol* 61.98 g/mol = 44.53 grams

There will be 44.53 grams of sodium oxide produced