Answer:
The first answer is 7,24g Mg
The second answer is 3,36L O2
Explanation:
Atomic masses: Mg= 24,3 ; O=16
Solution: MMgO = 40,3 g/mol
2 Mg + O2 => 2 MgO
1) 12 g MgO x (2x24,3 g Mg / 2x 40,3 MgO) = 7,24 g Mg
2)
At first calculate moles from Oxygen:
12 g MgO x ( 1 mol from Oxygen/ 2 x 40,3 g MgO ) = 0,15 moles from Oxygen
then ....
0,15 moles from Oxygen x (22,4 lts / 1 mol from Oxygen ) = 3,36 lts
22.4 liters is the constant that we use in this equation, it is a value that is repeated since we are in the presence of a gas with ideal behavior, so it is considered that the normal molar volume of ideal gaseous substances is always 22.4 liters, estimated with a temperature of 0º C and a pressure of 1 atmosphere. (constant temperature and pressure)
