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
