4.0 g Mg and 4.0 g O2 are placed in a container and magnesium oxide, MgO, forms. The Mg is totally consumed but 1.4 g O2 remains. How much magnesium oxide formed

The correct answer is: 6.6 g MgO

Explanation:

First we have to write and balance the chemical reaction as follows:

2Mg(s) + O₂(g) → 2MgO(s)

That means that 2 moles of Mg(s) react with 1 mol of O₂(g) to give 2 moles of MgO(s). If Mg is totally consumed and a mass of O₂ remains unaltered after reaction, the limiting reactant is Mg. We use the limiting reactant to calculate the mass of product.

According to the balanced chemical equation, 2 moles of Mg(s) produce 2 moles of MgO(s).

2 moles Mg = 2 mol x molar mas Mg= 2 mol  x 24.3 g/mol = 48.6 g Mg

2 moles MgO= 2 mol x (molar mass Mg + molar mass O) = 2 mol x (24.3 g/mol + 16 g/mol) = 80.6 g MgO

The stoichiometric ratio is 80.6 g MgO/48.6 g Mg. So, we multiply this ratio by the mass of consumed Mg (4.0 g) in order to obtain the produced mass of MgO:

4.0 g Mg x 80.6 g MgO/48.6 g Mg = 6.63 g MgO

6.6 grams of magnesium oxide are formed.

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i