What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP

Question

asked Aug 23, 2018 118k views

2 Answers

Afftee · Answer 1 · 2018-08-28T21:53:08+0000

Answer: 3808 ml

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
6.023* 10^(23) of particles.

Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=(4.03)/(24)=0.17moles$

$2Mg+O_2\rightarrow 2MgO$

2 moles of magnesium react with= [tex[2\times 22.4=44.8L[/tex] of oxygen at STP

Thus 0.17 moles of magnesium react with=
(44.8)/(2)* 0.17=3.8L=3808ml

Thus the volume of oxygen required to react with 4.03 g of magnesium at STP is 3808 ml.