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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?

Respuesta :

Answer:

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

joseaaronlara

Answer:

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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