In magnesium oxide (MgO), magnesium (Mg) is present as a cation with a +2 charge. The composition consists of one magnesium atom combined with one oxygen atom, resulting in a 1:1 molar ratio in the compound. The molar mass of magnesium is approximately 24.31 g/mol, while that of oxygen is about 16.00 g/mol. Thus, Mg contributes about 60.3% of the total mass of MgO.
60%
The reaction is:MgO + H2O = Mg(OH)2
The reaction is between Mg and O2. 2Mg+O2->2MgO
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
2 Mg (s) + O2 (g) → 2 MgO (s)4.00mg x 1 mol Mg/24.305 x 2 mol MgO/2 mol Mg x 40.305g MgO/1 mol MgO==6.63g4.00g- Given24.305g- Atomic Mass of Magnesium40.305g- Atomic Mass of Magnesium Oxide
The percent composition of magnesium (Mg) in MgO is 60%. This is because there is one magnesium atom (24.3 g/mol) and one oxygen atom (16.0 g/mol) in the formula unit MgO (40.3 g/mol), so dividing the molar mass of magnesium by the molar mass of MgO gives 0.603, or 60.3%.
The balanced equation for MgO + H2O is MgO + H2O -> Mg(OH)2.
The molar mass of MgO is 40.3 g/mol (Mg: 24.3 g/mol, O: 16 g/mol). The molar mass contribution of oxygen in MgO is 16 g/mol. Therefore, the percent composition of oxygen in MgO is (16 g/mol / 40.3 g/mol) * 100 = 39.7%.
60%
Magnesium nitride (Mg3N2) has a higher percentage composition of magnesium than magnesium oxide (MgO) does. The percentage composition of magnesium in magnesium nitride is 72.2% and the percentage composition of magnesium in magnesium oxide is 60.3%.
The percent of Mg calculated will be too high. Let's say that you reacted 1.00 g of Mg and made some MgO, but so much MgO escaped as smoke that only 1.00 g of MgO was left. You would then conclude from the numbers that the mass of O in the MgO was zero! This would lead you to conclude that the percent of Mg in the MgO was 100 %, which is silly and clearly in error. Although this is an extreme example, it illustrates that the loss of MgO as smoke from the crucible leads to a percent of Mg (calculated) that is above the expected 60.3 %.
"MgO" is magnesium oxide and "H" is hydrogen, as in "Mg + H(2)O => MgO + H(2)" MgO + H2 ---> H2O + Mg
mgo+h2o=mg(OH)2 AAHana
The reaction is:MgO + H2O = Mg(OH)2
The reaction is between Mg and O2. 2Mg+O2->2MgO
Yes! MgO+H2O=Mg((OH)2), Magnesium Hydroxide (a base).
60%