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.
To calculate the percent error of oxygen in magnesium oxide (MgO), you would compare the experimental value of oxygen in MgO to the theoretical value. The experimental value can be determined by chemical analysis, while the theoretical value can be calculated using the molecular formula of MgO. The percent error is calculated using the formula: (|Theoretical value - Experimental value| / Theoretical value) x 100%.
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.
magnesium oxide
The compound formed by magnesium and oxygen is called magnesium oxide.
Burning magnesium ribbon is a synthesis reaction because it involves the combination of magnesium with oxygen to form magnesium oxide. In this reaction, new chemical bonds are formed and a single product is produced from the combination of two reactants.
If completely burnt (in excess of oxygen), all the magnesium will be converted to magnesium oxide.
Magnesium oxide. It's produced by magnesium reacting with oxygen.
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The mass of heavy water produced when 7,00grams of oxygen reacts with excess D2 is 7,875 g.
To calculate the percent error of oxygen in magnesium oxide (MgO), you would compare the experimental value of oxygen in MgO to the theoretical value. The experimental value can be determined by chemical analysis, while the theoretical value can be calculated using the molecular formula of MgO. The percent error is calculated using the formula: (|Theoretical value - Experimental value| / Theoretical value) x 100%.
At the cathode, magnesium metal will be produced, while at the anode, oxygen gas will be produced. This is because during the electrolysis of magnesium sulfate, magnesium ions will be reduced at the cathode to form magnesium metal, and water molecules will be oxidized at the anode to form oxygen gas and hydrogen ions.
the white bright light produced in fire works is due to burning of magnesium in the presence of oxygen magnesium+oxygen=magnesiumoxide
Yes, the amount of magnesium ribbon burned does affect how much magnesium oxide is produced. More magnesium ribbon burned will result in more magnesium oxide being produced since the reaction between magnesium and oxygen is stoichiometric, meaning it requires a certain ratio of reactants to produce a set amount of 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.
If 3 grams of magnesium are used to form 4 grams of magnesium oxide, then 1 gram of oxygen is used in the reaction. This means 1 gram of oxygen remains unused.
If the magnesium is not polished, there may be impurities or oxides on the surface that could affect the reported mole ratio of oxygen to magnesium. This could result in a higher reported mole ratio due to the presence of excess oxygen-containing compounds on the surface, leading to an inaccurate measurement of the actual ratio of oxygen to magnesium.
When magnesium reacts with oxygen to form magnesium oxide, the ratio of magnesium to oxygen is fixed according to the chemical equation (2Mg + O2 -> 2MgO). This means that the amount of magnesium oxide that can be formed is limited by the amount of oxygen available, not by the amount of magnesium present. So, increasing the amount of magnesium does not lead to a corresponding increase in magnesium oxide produced.