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 mol O2x56.3g MgO2=1.24 g Mgo2;
0.041molMg-0.022molO2=0.019 mol Mg or 0.462g inexcess.
looks good to me, I didn't do the calculator punching.
I would personally have balanced the equation with whole numbers, but it looks ok.
See my response to your duplicate post above. Since you didn't balance the equation I think all of your work needs to be redone. For openers I don't believe O2 is the LR.
if 1.2.5 g of magnesium is ignited in a flask containg 2.500 liters of oxygen at STP which is the limiting reactant
what is the theoretical yield of Magnesium oxide
if the % yield is only 72.5% what is the actual yield of Magnesium oxide
To calculate the amount of magnesium oxide produced and identify the reactant in excess, you followed the correct steps. Let's break it down:
1. Determine the moles of magnesium (Mg):
Given that 1.00 g of magnesium represents a molar mass of 24.3 g/mol, you divided the mass (1.00 g) by the molar mass (24.3 g/mol). This yields 0.041 mol of Mg.
2. Determine the moles of oxygen (O2):
Given that the flask contains 0.500 liters of oxygen at STP, you used the ideal gas law to convert the volume to moles. This is done by dividing the volume (0.500 L) by the molar volume at STP (22.4 L/mol). This gives you 0.022 mol of O2.
3. Identify the limiting reactant (LR):
Since the stoichiometry of the balanced equation tells us that the ratio of moles between Mg and O2 is 1:1, it is important to compare the actual ratio of moles. In this case, the amount of O2 (0.022 mol) is smaller than the amount of Mg (0.041 mol). Therefore, O2 is the limiting reactant.
4. Calculate the mass of magnesium oxide (MgO) formed:
Use the stoichiometry from the balanced equation: 1 mol of O2 reacts with 2 mol of Mg to form 2 mol of MgO. Thus, 0.022 mol of O2 would react with 0.022 mol of Mg, forming 0.022 mol of MgO. To get the mass, multiply the number of moles of MgO by its molar mass (56.3 g/mol). This yields 1.24 g of MgO.
5. Calculate the amount of reactant in excess:
To find the amount of Mg that is in excess, subtract the number of moles of O2 used (0.022 mol) from the moles of Mg initially present (0.041 mol). This gives you 0.019 mol of excess Mg. To convert this to grams, multiply by the molar mass of Mg (24.3 g/mol), resulting in 0.462 g of excess Mg.
So, in summary, 1.24 grams of magnesium oxide (MgO) are produced, and the reactant in excess is magnesium (Mg), with 0.462 grams remaining.