I'm not looking for an answer, just how to find it. Mass of crucible and cover: 37.47 g; Mass of crucible, cover, and magnesium: 37.92 g; Mass of crucible, cover, and product 1st time (before adding water): 30.18 g; Mass of crucible, cover, and product 2nd time: 38.15 g.
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First you want the mass of magnesium, which is 37.92 - 37.47 = 0.45 g.
You shouldn't have lost mass to the atmosphere in this reaction though. (Looks like you burned up part of the crucible, which is really strange). I suspect therefore that 30.18 is a typo. Perhaps you meant 38.18 g, which is reasonable. In which case, subtract the 37.47 g and you get 0.71 g of product, which may or may not be pure MgO.
If you got 100% yield then 0.45 g Mg / (24.3050 g/mol) = 0.0185 mol Mg, which produces 0.0185 mol MgO, which * (40.3044 g/mol) = 0.746 g MgO theoretical.
How much of the Mg really reacted? x g Mg produces (40.3044 / 24.3050) x g of MgO, using the same reasoning we gave above, where x is the amount that actually reacted. We start with 0.45 g Mg and we end up with (40.3044 / 24.3050) x g of MgO *and* 0.45 - x g of unreacted Mg. We calculate 40.3044 / 24.3050 = 1.658, and we reason that 1.658 x g MgO + (0.45 - x) g Mg have a total mass of 0.71 g. From this point it's a matter of algebra:
1.658 x + (0.45 - x) = 0.71
(1.658 - 1) x = 0.71 - 0.45
0.658 x = 0.26
x = 0.40 g of Mg which reacted
0.45 - x = 0.05 g of unreacted Mg
1.658 x = 0.66 g of MgO.
And there's your answer.
You shouldn't have lost mass to the atmosphere in this reaction though. (Looks like you burned up part of the crucible, which is really strange). I suspect therefore that 30.18 is a typo. Perhaps you meant 38.18 g, which is reasonable. In which case, subtract the 37.47 g and you get 0.71 g of product, which may or may not be pure MgO.
If you got 100% yield then 0.45 g Mg / (24.3050 g/mol) = 0.0185 mol Mg, which produces 0.0185 mol MgO, which * (40.3044 g/mol) = 0.746 g MgO theoretical.
How much of the Mg really reacted? x g Mg produces (40.3044 / 24.3050) x g of MgO, using the same reasoning we gave above, where x is the amount that actually reacted. We start with 0.45 g Mg and we end up with (40.3044 / 24.3050) x g of MgO *and* 0.45 - x g of unreacted Mg. We calculate 40.3044 / 24.3050 = 1.658, and we reason that 1.658 x g MgO + (0.45 - x) g Mg have a total mass of 0.71 g. From this point it's a matter of algebra:
1.658 x + (0.45 - x) = 0.71
(1.658 - 1) x = 0.71 - 0.45
0.658 x = 0.26
x = 0.40 g of Mg which reacted
0.45 - x = 0.05 g of unreacted Mg
1.658 x = 0.66 g of MgO.
And there's your answer.