Question

how many gram of O2 is required for the combustion 480 grams of Mg? find the maass of MgO formed in this reaction .​

Answer 1

Answer: 320 gram of $O_2$ is required for the combustion of 480 grams of Mg.

The mass of $MgO$  formed in this reaction is 800 g.

Explanation:

$2Mg+O_2\rightarrow 2MgO$

According to avogadro's law, 1 mole of every substance contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

$\text{Number of moles}=\frac{480g}{24g/mol}=20moles$

Now 2 moles of $Mg$ reacts with 1 mole of $O_2$

20 moles of $Mg$ reacts with=$\frac{1}{2}\times 20=10$ moles of $O_2$

Mass of $O_2=\{\text{number of moles}}\times {\text{Molar mass}}=10\times 32=320 g$

2) Now 2 moles of $Mg$ produces 2 moles of $MgO$

20 moles of $Mg$ reacts with=$\frac{2}{2}\times 20=20$ moles of $MgO$

Mass of MgO=$\{\text{number of moles}}\times {\text{Molar mass}}=20\times 40=800g$

Answer 2

given, mass of Mg = 480 grams

so, mole of Mg = 480/24 = 20

combustion of magnesium is shown by chemical reaction ..

$2Mg+O_2\rightarrow 2MgO2$

in this reaction, it is clear that 1 mole of O2 is required for combustion for 2 moles of Mg.

so, 10 moles of O2 is required for combustion for 20 moles of Mg.

hence, weight of O2 is required = mole of O2 × molecular weight

= 10 × 32 = 320 grams

also we see that, 2 mole of Mg formed 2 mole of MgO

so, 20 moles of Mg formed 20 moles of MgO

hence, weight of MgO = number of mole of MgO × molecular weight

= 20 × (24 + 16)

= 20 × 40 = 800 grams.