To determine how many moles of Mercury (II) oxide (HgO) are needed to produce 125 g of oxygen (O₂), we first need to consider the decomposition reaction: 2 HgO(s) → 2 Hg(l) + O₂(g). From this equation, we see that 2 moles of HgO produce 1 mole of O₂. The molar mass of O₂ is approximately 32 g/mol, so 125 g of O₂ corresponds to about 3.91 moles (125 g ÷ 32 g/mol). Therefore, since 2 moles of HgO produce 1 mole of O₂, we need 7.82 moles of HgO (3.91 moles O₂ × 2 moles HgO/mole O₂).
800 g oxygen are needed.
Four moles of potassium chlorate are needed.
To find the moles of cesium chlorate needed to produce 2.7 moles of oxygen gas, use the balanced chemical equation for the decomposition of cesium chlorate: 2CsClO3 -> 2CsCl + 3O2 From the equation, it shows that 2 moles of cesium chlorate produce 3 moles of oxygen gas. Therefore, you will need (2/3) * 2.7 = 1.8 moles of cesium chlorate to produce 2.7 moles of oxygen gas.
8,75 moles of oxygen are needed.
To determine how many moles of mercury are produced when 125 g of oxygen is generated, we first need to know the balanced chemical equation for the reaction involving mercury and oxygen. Assuming the reaction is the formation of mercury(II) oxide (HgO) from mercury (Hg) and oxygen (O₂), the equation is: 2 Hg + O₂ → 2 HgO. Given that the molar mass of oxygen (O₂) is approximately 32 g/mol, 125 g of oxygen corresponds to about 3.91 moles of O₂. According to the stoichiometry of the balanced equation, 2 moles of Hg are produced for every 1 mole of O₂. Therefore, 3.91 moles of O₂ would produce approximately 7.82 moles of Hg.
800 g oxygen are needed.
The chemical reactin is:2 KClO3 = 2 KCl + 3 O24 moles of potassium chlorate produce 6 moles oxygen.
Four moles of potassium chlorate are needed.
For the decomposition of potassium chlorate, the molar ratio between potassium chlorate (KClO3) and oxygen (O2) is 2:3. Therefore, to produce 15 moles of oxygen, 10 moles of potassium chlorate are needed. (15 moles O2) x (2 moles KClO3 / 3 moles O2) = 10 moles KClO3.
To find the moles of cesium chlorate needed to produce 2.7 moles of oxygen gas, use the balanced chemical equation for the decomposition of cesium chlorate: 2CsClO3 -> 2CsCl + 3O2 From the equation, it shows that 2 moles of cesium chlorate produce 3 moles of oxygen gas. Therefore, you will need (2/3) * 2.7 = 1.8 moles of cesium chlorate to produce 2.7 moles of oxygen gas.
8,75 moles of oxygen are needed.
To determine how many moles of mercury are produced when 125 g of oxygen is generated, we first need to know the balanced chemical equation for the reaction involving mercury and oxygen. Assuming the reaction is the formation of mercury(II) oxide (HgO) from mercury (Hg) and oxygen (O₂), the equation is: 2 Hg + O₂ → 2 HgO. Given that the molar mass of oxygen (O₂) is approximately 32 g/mol, 125 g of oxygen corresponds to about 3.91 moles of O₂. According to the stoichiometry of the balanced equation, 2 moles of Hg are produced for every 1 mole of O₂. Therefore, 3.91 moles of O₂ would produce approximately 7.82 moles of Hg.
The balanced chemical equation for the reaction between iron and oxygen to produce Fe2O3 is 4Fe + 3O2 -> 2Fe2O3. From the equation, we see that 3 moles of oxygen react with 4 moles of iron to produce 2 moles of Fe2O3. Therefore, to find the grams of oxygen needed, we need to calculate the molar mass of Fe2O3 and then determine the number of grams needed using the mole ratio from the balanced equation.
16,875 moles of oxygen are needed.
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
One mole of sulfur reacts with 1.5 moles of oxygen to produce one mole of sulfur trioxide. So, with two moles of sulfur and three moles of oxygen, the limiting reactant is sulfur. Therefore, two moles of sulfur will produce two moles of sulfur trioxide.
When 2.5 moles of oxygen react with hydrogen, they react in a 1:2 ratio to produce water. Therefore, 2.5 moles of oxygen will produce 5 moles of water. To convert moles to grams, you'll need to know the molar mass of water, which is approximately 18 grams/mol. So, 2.5 moles of oxygen will produce 90 grams (5 moles x 18 grams/mole) of water.