Question

Se ha quemado magnesio (reacción con el oxígeno) y se obtuvieron 12g de óxido de magnesio (II). ¿Cuánto magnesio se quemó? ¿Qué volumen de oxígeno medido en condiciones normales se quemó?

Answer 1

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)