if 250 ml of oxygen at stp are consumed when magnesium burns, how many grams of magnesium oxide? [Mg(s) + O2(g) ---> MgO(s)]
How many mols O2 were consumed? That's 0.250 L x (1 mol/22.4L) = ? mols O2.
Use the coefficients in the balanced equation to convert mols O2 to mols MgO.
Now convert mols MgO to g. g = mols x molar mass.
i can't find the answer can u show me how to do the work step by step and tell me the answer my homework is due tomorrow
To determine the number of grams of magnesium oxide (MgO) produced when 250 ml of oxygen gas (O2) at standard temperature and pressure (STP) is consumed during the burning of magnesium (Mg), we need to follow these steps:
1. Calculate the number of moles of oxygen gas (O2) using the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. At STP, the pressure is 1 atm, the volume is 22.4 L per mole, and the temperature is 273 Kelvin.
Since we have the volume in milliliters (ml), we need to convert it to liters (L):
250 ml = 250/1000 L = 0.25 L
So, the number of moles of oxygen gas (O2) can be calculated as:
n = (P * V) / (R * T)
= (1 atm * 0.25 L) / (0.0821 L * atm/mol * K * 273 K)
≈ 0.010 mol
2. Based on the balanced chemical equation for the reaction (Mg + O2 ---> MgO), we can see that 1 mol of magnesium (Mg) reacts with 1 mol of oxygen (O2) to produce 1 mol of magnesium oxide (MgO).
Therefore, the number of moles of magnesium oxide (MgO) produced is also 0.010 mol.
3. To find the mass of magnesium oxide (MgO) produced, we need to know the molar mass of MgO. The atomic masses of magnesium (Mg) and oxygen (O) are approximately 24.31 g/mol and 16.00 g/mol, respectively.
The molar mass of magnesium oxide (MgO) is:
Molar mass of MgO = (1 * molar mass of Mg) + (1 * molar mass of O)
= (1 * 24.31 g/mol) + (1 * 16.00 g/mol)
≈ 40.31 g/mol
4. Finally, we can calculate the mass of magnesium oxide (MgO):
Mass of MgO = Number of moles of MgO * Molar mass of MgO
= 0.010 mol * 40.31 g/mol
= 0.4031 g
Therefore, approximately 0.4031 grams of magnesium oxide (MgO) will be produced when 250 ml of oxygen at STP are consumed during the burning of magnesium.