The answer is 3,375 moles oxygen.
We need 3 moles of potassium perchlorate.
Four moles of potassium chlorate are needed.
Oxygen limits the reaction, so......Balanced equation. 2H2 + O2 -> 2H2O 7.89 mole H2O (1 mole O2/2 mole H2O) = 3.95 mole oxygen gas needed ------------------------------------------
To determine how many grams of mercury (Hg) are needed to produce 3.75 moles of oxygen (O₂), we first consider the balanced chemical reaction for the formation of oxygen from mercury(II) oxide (HgO): 2 HgO → 2 Hg + O₂. From this, we see that 2 moles of HgO produce 1 mole of O₂, which means 3.75 moles of O₂ require 7.5 moles of Hg. The molar mass of Hg is approximately 200.59 g/mol, so 7.5 moles of Hg would weigh about 1,504.43 grams (7.5 moles × 200.59 g/mol).
24 mols of aluminum must have reacted since aluminum oxide is Al2O3
Sounds like an acid metal reaction to produce hydrogen gas, so I will assume you meant hydrochloric acid. Complete reaction indicates aluminum is limiting and drives the reaction. 2Al + 6HCl --> 2AlCl3 + 3H2 3.0 moles Al (3 moles H2/2 moles Al) = 4.5 mole hydrogen gas produced =========================
Four moles of potassium chlorate are needed.
We need 3 moles of potassium perchlorate.
It takes 2 moles of NO to form 2 moles of NO2, so to form 4.67 moles of NO2 you would need 4.67 moles of NO.
Oxygen limits the reaction, so......Balanced equation. 2H2 + O2 -> 2H2O 7.89 mole H2O (1 mole O2/2 mole H2O) = 3.95 mole oxygen gas needed ------------------------------------------
2Cl2 + 7O2 --> 2Cl2O7So for every 2 moles of dichlorine heptoxide formed, 7 moles of oxygen are reacted.69.1 moles Cl2O7 * (7 moles O2 / 2 moles Cl2O7) = 241.85 moles O2
2KClO3==>2KCl+3O2 is the equation. so you need 4 moles of KClO3.
Sulfur dioxide is SO2, so there are 2 moles of oxygen for each mole of sulfur. In 25 moles of SO2, there are 25 moles of S, so there must be 50 moles of oxygen, or 25 moles of diatomic oxygen gas (O2).
To determine how many grams of mercury (Hg) are needed to produce 3.75 moles of oxygen (O₂), we first consider the balanced chemical reaction for the formation of oxygen from mercury(II) oxide (HgO): 2 HgO → 2 Hg + O₂. From this, we see that 2 moles of HgO produce 1 mole of O₂, which means 3.75 moles of O₂ require 7.5 moles of Hg. The molar mass of Hg is approximately 200.59 g/mol, so 7.5 moles of Hg would weigh about 1,504.43 grams (7.5 moles × 200.59 g/mol).
2KClO3 --> 2KCl + 3O2For every 3 moles of oxygen gas produced, 2 moles of potassium chlorate are used.6 moles O2 * (2 moles KClO3 reacted / 3 moles O2 produced) = 4 moles KClO3
24 mols of aluminum must have reacted since aluminum oxide is Al2O3
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.