If 84 grams of sodium chloride reacts with an excess amount of magnesium oxide, how many grams of sodium oxide will be produced? question 2 options: 23.2 g na20 45g mgcl2 107g mgo 44.5g na20

44.5 g of Na₂O

Explanation:

The reaction is this one:

2NaCl + MgO  →  Na₂O  +  MgCl₂

Moles of NaCl = Mass / Molar mass

84 g / 58.45 g/m = 1.43 moles

Ratio is 2:1, so if we produce 1 mol of Na₂O, from 2 moles of NaCl; If we have 1.43 moles, we 'll produce the half of moles

1.43 / 2 = 0.72 moles

Molar mass Na₂O = 62 g/m

Mol . molar mass =  0.72 m . 62 g/m = 44.5 g

There will be 44.5 grams of sodium oxide (Na2O) produced

Explanation:

Step 1: Data given

Mass of Sodium chloride (NaCl) = 84.00 grams

Magnesium oxide = in excess

Molar mass of NaCl = 58.44 g/mol

Molar mass of sod)ium oxide (Na2O = 61.98 g/mol

Step 2: The balanced equation

2NaCl + MgO → Na2O + MgCl2

Step 3: Calculate moles of NaCl

Moles NaCl = Mass / Molar mass

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 NaCl consumed, we need 1 mol MgO to produce 1 mol 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 Na2O = Moles Na2O * Molar mass Na2O

Mass Na2O = 0.7185 moles * 61.98 g/mol

Mass Na2O = 44.53 grams of Na2O

There will be 44.5 grams of sodium oxide (Na2O) produced

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