2nacl + mgo > na20 + mgcl2

if 84 grams of sodium chloride reacts with an excess amount of magnesium oxide, how many grams of sodium oxide will be produced?

There will be 44.53 grams of sodium oxide produced

Explanation:

Step 1: Data given

Mass of sodium chloride = 84.00 grams ( = the limiting reactant)

Molar mass of NaCl = 58.44 g/mol

Magnesium oxide is in excess

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

The answer to your question is 44.13 g of Na₂O.

Explanation:

Chemical equation

                               2NaCl   +   MgO  ⇒  Na₂O  +  MgCl₂

Data                         84 g                              ?

Process

1.- Calculate the molecular weight of NaCl and Na₂O.

NaCl = 23 + 36 = 59 g

Na₂O = (23 x 2) + 16 = 62 g

2.- Solve this problem using proportions

                          2(59)g of NaCl 62 g of Na₂O

                          84 g of NaCl      x

                          x = (84 x 62) / 2(59)

                          x = 5208 / 118

                         x = 44.13 g of Na₂O                  

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