what mass of magnesium is needed to reduce a 500ml sample of oxygen at stp to a 375ml sample?
Mg + O2--> MgO + O2
To answer this question, we need to use stoichiometry - the relationship between the balanced equation and the quantities of reactants and products involved.
First, we need to determine the mole ratio between magnesium (Mg) and oxygen (O2) in the balanced equation:
Mg + O2 -> MgO + O2
From the equation, we can see that 1 mole of magnesium reacts with 1 mole of oxygen. This means that the mole ratio of Mg to O2 is 1:1.
Next, we need to calculate the number of moles of oxygen in both the initial and final samples using the ideal gas law. At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.
The initial volume of oxygen (V1) is given as 500 ml. To convert it to liters, we divide by 1000:
V1 = 500 ml / 1000 ml/L = 0.5 L
Similarly, the final volume of oxygen (V2) is given as 375 ml:
V2 = 375 ml / 1000 ml/L = 0.375 L
Now, we can calculate the number of moles of oxygen in each sample:
n1 = V1 / V_Molar
where V_Molar is the molar volume at STP, which is 22.4 L/mol.
n1 = 0.5 L / 22.4 L/mol = 0.02232 mol
n2 = V2 / V_Molar
n2 = 0.375 L / 22.4 L/mol = 0.01674 mol
The oxygen has been reduced from 0.02232 mol to 0.01674 mol.
Since the mole ratio of Mg to O2 is 1:1, we can conclude that 0.02232 moles of magnesium were needed to reduce 0.02232 moles of oxygen.
To calculate the mass of magnesium needed, we'll need the molar mass of magnesium (24.31 g/mol).
Mass = moles × molar mass
Mass = 0.02232 mol × 24.31 g/mol = 0.541 g
Therefore, approximately 0.541 grams of magnesium is needed to reduce the 500 ml sample of oxygen at STP to a 375 ml sample.