You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
Why did you post this again? Because you're lazy and wanted people to not see my post? Once again, ONE QUESTION AT A TIME. We're here to help you with a tough question or work you through it, not do all your homework.
Answer:
because they devlop our organs
The final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L.
<h3>How to calculate volume?</h3>
The volume of a given gas can be calculated using the Charles law equation as follows:
V1/T1 = V2/T2
Where;
- V1 = initial volume
- V2 = final volume
- T1 = initial temperature
- T2 = final temperature
- V1 = 2L
- V2 = ?
- T1 = -25°C + 273 = 248K
- T2 = 273K
2/248 = V2/273
273 × 2 = 248V2
546 = 248V2
V2 = 546/248
V2 = 2.2L
Therefore, the final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L
Learn more about volume at: brainly.com/question/11464844
