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

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.