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 [tex]6.023\times 10^{23}[/tex] 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:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{4.03}{24}=0.17moles[/tex]
[tex]2Mg+O_2\rightarrow 2MgO[/tex]
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=[tex]\frac{44.8}{2}\times 0.17=3.8L=3808ml[/tex]
Thus the volume of oxygen required to react with 4.03 g of magnesium at STP is 3808 ml.
