How many moles of oxygen are required for the combustion of 35.5 g of Magnesium Mg+O2-->MgO Please balance

0.74 moles of oxygen are required for combustion

Explanation:

Given data:

Number of moles of oxygen required = ?

Mass of Mg = 35.5 g

Solution:

Chemical equation:

2Mg + O₂         →       2MgO

Number of moles of Mg:

Number of moles = mass/molar mass

Number of moles = 35.5 g/ 24 g/mol

Number of moles = 1.48 mol

Now we will compare the moles of Mg with oxygen.

                  Mg           :         O₂

                   2             :          1

                 1.48           :        1/2×1.48 = 0.74 mol

0.74 moles of oxygen are required for combustion.

indium, and thallium have three electrons i think

they are all solids.