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(106 g MgO / 1) * ( 1 mol MgO / 40.3044 g MgO) = 2.63 mol MgO.
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How many moles are present in 160g of MgO?
40.3 g of MgO = 1 mole So 160 g of MgO = (1 *160) / 40.3 = 3.97 mole
How many grams of MgO are produced when 40 grams of 02 react with Mg?
:Mg: (2.43 g)/(24.3 g/mol) = .1 mol :MgO: (.1 mol)(24.3+16.0 g/mol) = 4.03 g
How many moles of so3 are in 20g of so3?
The molar mass of sulfur dioxide is 64,066 g.
What theoretical mass of magnesium oxide is produced when 82.56 grams of magnesium reacts with excess oxygen?
First write the balanced chemical equation: 2 Mg + O2 --> 2 MgO. Now solve for the theoretical amount of MgO formed: (82.56 g Mg / 1) * (1 mol Mg / 24.305 g Mg) * (2 mol MgO / 2 mol Mg) * (40.3044 g MgO / 1 mol MgO) = 136.9 g MgO.
Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product what would be the theoretical yield of the product?
2Mg + O2 --> 2MgO First, take the amount given (2.033g Mg) and convert it to moles. 1 mole of Mg weighs 24.312g. Then, convert to the desired element, which is MgO. Then convert the moles of MgO into grams. The complete problem looks like this: 2.033g Mg(1 mol Mg/24.312g Mg)(2 mol MgO/2 mol Mg)(40.311g MgO/1 mol MgO) = 3.371 g MgO 3.371 g MgO should be produced.
How many moles are in 106 grams of MgO?
(106 g MgO / 1) * ( 1 mol MgO / 40.3044 g MgO) = 2.63 mol MgO.
How many moles are present in 160g of MgO?
40.3 g of MgO = 1 mole So 160 g of MgO = (1 *160) / 40.3 = 3.97 mole
What is the mass of magnesium oxide when you burn 4 grams of magnesium?
Mg grams -> (use Mg's molar mass) -> Mg moles -> (use ratio of moles - use balanced equation) -> MgO moles -> (use MgO's molar mass) -> grams MgO set up the equation: Mg + O2 --> MgO (we know the product is MgO and not MgO2 because magnesium has a charge of 2+ while oxygen has a charge or 2-) balance the equation: 2Mg + O2 --> 2MgO Molar mass of Mg: 24.31 g/mol Molar mass of MgO: 44.30 g/mol (add the molar mass of Magnesium - 24.31g/mol and the molar mass of Oxygen - 15.99g/mol together) (use periodic table to find these) 7.0 grams of Mg To find the moles of Magnesium you use the molar mass of Mg. (7.0 g Mg)*(1 mol Mg / 24.31 g Mg) =0.2879 moles Mg notice how the grams cancel to leave you with moles - remember dividing by a fraction is the same as multiplying by the reciprocal Now use the balanced equation's coefficients and the moles of Mg to determine the number of moles of MgO present. 2Mg + O2 --> 2MgO 2 moles Mg : 2 moles MgO -> divide both sides by 2 and it obviously becomes a 'one to one' ratio. This means that the number of moles of Mg is equal to the number of moles of MgO. This means that there are 0.2879 moles of MgO. Now that we know MgO's molar mass and the number of moles of MgO we have, the grams of MgO produced can be determined. (0.2879 moles MgO)*(44.30 g MgO / 1 mol MgO) = 12.75 grams MgO
How many grams of magnesium and oxygen react to form 20 g of magnesium oxide?
Balanced Formula:2Mg + O2 --> 2MgOMole ratio:2 : 1 : 2Givens:.486 g oxygen.738 g magnesium24.3 g = atomic mass of magnesium16.0 g = atomic mass of oxygen40.3 g = molecular mass of magnesium oxideFind the amount (in moles) of Magnesium oxide that oneelement will make:(.486 g O) / (16.0 g O) × (2 moles MgO)= .0608 moles MgO(.783 g Mg) / (24.3 g Mg) = .0322 moles MgOThere is less MgO produced with magnesium than oxygen; therefore, magnesium is the limiting reactant and the oxygen is the excess reactant. The magnesium determines how much Magnesium oxide is produced. It would be good to get .0608 moles of MgO, but there isn't enough magnesium. So the amount of MgO produced will be determined on the amount of Magnesium.Convert moles of MgO produced with the amount of oxygen to grams:.0322 mol MgO (40.3 g) = 1.30 grams of MgO produced--------------------------------------------------------------------------------------------------------You will need 3 moles of oxygen if you start with six moles of magnesium. This will allow you to produce 6 moles of magnesium oxide.Source: (e2020)
When 3.00 g of Mg is ignited in 2.20 g of pure oxygen what is the limiting reactant and what is the theoretical yield of MgO?
Moles Mg = 3.00 g / 24.312 g/mol =0.123 Moles O2 = 2.20 / 32 g/mol = 0.0688 2 Mg + O2 >> 2 MgO the ratio between Mg and O2 is 2 : 1 0.123 / 2 = 0.0615 moles O2 needed we have 0.0688 moles of O2 so O2 is in excess and Mg is the limiting reactant we get 0.123 moles of MgO => 0.123 mol x 40.31 g/mol =4.96 g
How many grams of MgO are produced when 40 grams of 02 react with Mg?
:Mg: (2.43 g)/(24.3 g/mol) = .1 mol :MgO: (.1 mol)(24.3+16.0 g/mol) = 4.03 g
How many moles of so3 are in 20g of so3?
The molar mass of sulfur dioxide is 64,066 g.
What theoretical mass of magnesium oxide is produced when 82.56 grams of magnesium reacts with excess oxygen?
First write the balanced chemical equation: 2 Mg + O2 --> 2 MgO. Now solve for the theoretical amount of MgO formed: (82.56 g Mg / 1) * (1 mol Mg / 24.305 g Mg) * (2 mol MgO / 2 mol Mg) * (40.3044 g MgO / 1 mol MgO) = 136.9 g MgO.
Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product what would be the theoretical yield of the product?
2Mg + O2 --> 2MgO First, take the amount given (2.033g Mg) and convert it to moles. 1 mole of Mg weighs 24.312g. Then, convert to the desired element, which is MgO. Then convert the moles of MgO into grams. The complete problem looks like this: 2.033g Mg(1 mol Mg/24.312g Mg)(2 mol MgO/2 mol Mg)(40.311g MgO/1 mol MgO) = 3.371 g MgO 3.371 g MgO should be produced.
How many moles does 3.80 g Zn have?
3,80 g Zn have 0,058 moles.
If you have g of beryllium how many moles is this equivalent to?
The formula is: number of moles = g Be/9,012.
How many moles of argon are in 24.7 g of argon?
Moles = weight in g / atomic weight. So moles in 24.7 g of Ar = 24.7 / 39.948 = 0.62 moles
