Mg + O2 ------> MgO
if 1.00 g of magnesium is ignited in a flask containing 0.500liters of oxygen at STP, how many grams of magnesium oxide areproduced?
what is the name and the amount of the reactant inexcess?
My work:
1g Mg/24.3g=0.041 mol Mg;
0.5L O2/22.4L=0.022 mol O2; So O2 is the LR; 0.022 molx 56.3g/mol=1.24g MgO2.
0.041mol Mg-0.022mol O2=1.24mol Mg inexcess. Am I right? Thanks!
No. You didn't balance the equation.
2Mg + O2 ==> 2MgO
I'll let you redo it form here.
Thanks for posting your work. It makes it easy to check that way.
Reost here if you want me to recheck.
Thanks a lot!
Here's original question:
If 1.00 of magnesium is ignited in a flask containing 0.500 liter of oxygen at STP, how many grams of magnesium oxide are produced?
My work:
Mg + O2 ------> MgO2, is this right?
0.041mol Mg-0.022mol O2=0.019mol Mg inexcess. Thank you!
Lisa, you have to balance equations with the right compound formulas. 2Mg+O2>>2MgO
moles O2:.5/22.4 =.022moles
moles Mg:Magnesium oxide is MgO 0.041 mol Mg;
but you need twice as many moles of Mg as O2 (see balanced equation), so you need then .0205 moles O2. You have more than that, so the LR is Mg.
how many moles of MgO are produced: same number of moles of Mg: .041
recheck my numbers.
2Mg + O2 ==> 2MgO
1g Mg/24.3g=0.041 molMg, 0.041mol/2mol=0.0205 mol Mg;
0.5L O2/22.4L=0.0223 molO2; So Mg is the LR(0.0205mol Mg<0.0223 molO2);
0.041 molx40.3g MgO/mol=1.65g MgO2.
0.022molO2-0.0205molMg =0.0018 mol O2 inexcess.
Thanks!
Your calculations are mostly correct, but there is a small mistake in your final answer. Let's go through the calculations step by step to determine the correct answer:
1. Calculate the number of moles of magnesium (Mg):
Given: 1.00 g Mg, molar mass of Mg = 24.3 g/mol
Number of moles of Mg = 1.00 g Mg / 24.3 g/mol = 0.041 mol Mg
2. Calculate the number of moles of oxygen (O2):
Given: 0.500 L O2 at STP, molar volume of gas at STP = 22.4 L/mol
Number of moles of O2 = 0.500 L O2 / 22.4 L/mol = 0.022 mol O2
3. Determine the limiting reactant:
The balanced chemical equation shows that the ratio of moles of Mg to moles of O2 is 2:1. Comparing the moles calculated above, Mg has a lower mole value, indicating that it is the limiting reactant (LR).
4. Calculate the mass of magnesium oxide (MgO) produced:
Using the stoichiometry ratio from the balanced equation, 2 moles of Mg produce 2 moles of MgO.
Number of moles of MgO produced = 0.022 mol O2 (since it is the LR) x (2 mol MgO / 2 mol O2) = 0.022 mol MgO
Mass of MgO produced = 0.022 mol MgO x 40.3 g/mol (molar mass of MgO) = 0.888 g MgO
So, the correct answer is that 0.888 grams of magnesium oxide are produced.
Regarding the excess reactant, you calculated that there is 1.24 moles of Mg remaining. However, there seems to be a mistake in the formula you used. It should be Mg instead of MgO2. Additionally, the unit should be grams, not moles. Let's correct that:
The amount of excess reactant (Mg) can be calculated as follows:
Number of moles of excess Mg = 0.041 mol Mg - 0.022 mol O2 (since O2 is the LR) = 0.019 mol Mg
Mass of excess Mg = 0.019 mol Mg x 24.3 g/mol (molar mass of Mg) = 0.462 g Mg
Therefore, there is 0.462 grams of excess magnesium (Mg) remaining.
I hope this clarifies the calculations for you! Let me know if you have any further questions.