Question

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

Answer 1

Vol.250 before its to much pressure

Answer 2

Answer: 3808 ml

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ 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:

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

$2Mg+O_2\rightarrow 2MgO$

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=$\frac{44.8}{2}\times 0.17=3.8L=3808ml$

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