:Mg: (2.43 g)/(24.3 g/mol) = .1 mol :MgO: (.1 mol)(24.3+16.0 g/mol) = 4.03 g
To determine the number of moles of MgO produced from 11.2 L of O2, you would first need to balance the chemical equation for the reaction involving MgO and O2. Then, using the ideal gas law and stoichiometry, you can calculate the moles of MgO produced.
To find the number of moles in 160g of MgO, you first need to calculate the molar mass of MgO which is 40.3 g/mol for Mg + 16.0 g/mol for O = 56.3 g/mol for MgO. Then, divide the given mass by the molar mass to get the number of moles: 160g / 56.3 g/mol = 2.84 moles of MgO.
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.
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.
No. MgO is a base. How is it a base you say? Look at the following chemical equation: MgO + H20 ----> MgOH + OH- And then, MgOH -----> Mg+ + OH- So I guess you get two for one. I'm only a chemistry student, so you might want to look somewhere else too for an answer.
To determine the number of moles of MgO produced from 11.2 L of O2, you would first need to balance the chemical equation for the reaction involving MgO and O2. Then, using the ideal gas law and stoichiometry, you can calculate the moles of MgO produced.
To determine the number of moles in 106 grams of MgO, you need to divide the given mass by the molar mass of MgO. The molar mass of MgO is 40.3 g/mol (24.3 g/mol for Mg + 16 g/mol for O). Thus, 106 g / 40.3 g/mol = approximately 2.63 moles of MgO.
To find the number of moles in 160g of MgO, you first need to calculate the molar mass of MgO which is 40.3 g/mol for Mg + 16.0 g/mol for O = 56.3 g/mol for MgO. Then, divide the given mass by the molar mass to get the number of moles: 160g / 56.3 g/mol = 2.84 moles of MgO.
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.
magnesium oxide (MgO)
The balanced chemical equation for the reaction is: 2 Mg + O2 → 2 MgO. This means that for every 1 mole of O2, 2 moles of MgO are produced. Therefore, if 0.200 mol of O2 reacts completely, it will produce 0.400 mol of MgO.
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)
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
2Mg + O2 -----> 2MgO So two moles of magnesium oxide are formed if x moles of magnesium are allowed to react with only 1 mole of oxygen molecules. The oxygen has become the limiting ingredient.
there are two atoms in the chemical MgO
Magnesium oxide (MgO) is not attacked by atmospheric oxygen because MgO has a high heat of formation and a stable lattice structure. This makes it energetically unfavorable for oxygen to react with MgO under normal atmospheric conditions.
Yes, they react to produce ammonia and magnesium oxide. Mg3N2 + 3H2O --> 3 MgO + 2NH3