what mass of magnesium is needed to reduce a 500ml sample of oxygen at stp to a 375ml sample?
Mg + O2--> MgO + O2
To determine the mass of magnesium needed to reduce a 500 mL sample of oxygen to a 375 mL sample, we need to use stoichiometry and the balanced chemical equation provided.
1. Write down the balanced equation:
Mg + O2 --> MgO + O2
2. Determine the mole ratio:
From the equation, we can see that 1 mole of magnesium (Mg) reacts with 1 mole of oxygen (O2) to produce 1 mole of magnesium oxide (MgO).
3. Convert the volume of oxygen to moles:
The given volumes of the oxygen samples are at STP (Standard Temperature and Pressure), which means the molar volume is 22.4 L/mol. To convert the volumes to moles, we use the relationship:
22.4 L = 1 mole
For the 500 mL sample:
500 mL / 1000 mL = 0.5 L
0.5 L / 22.4 L/mol ≈ 0.0223 moles of O2
For the 375 mL sample:
375 mL / 1000 mL = 0.375 L
0.375 L / 22.4 L/mol ≈ 0.0167 moles of O2
4. Determine the moles of magnesium required:
Since the mole ratio is 1:1 between magnesium and oxygen, the moles of magnesium needed will be the same as the moles of oxygen.
Moles of Mg = Moles of O2 ≈ 0.0167 moles
5. Convert moles of magnesium to mass:
The molar mass of magnesium (Mg) is approximately 24.31 g/mol. To convert moles to grams, we use the formula:
Mass (g) = Moles × Molar mass
Mass of Mg = 0.0167 moles × 24.31 g/mol ≈ 0.405 g
Therefore, approximately 0.405 grams of magnesium is needed to reduce a 500 mL sample of oxygen at STP to a 375 mL sample.