3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O is the equation if it is dilute nitric acid. In concentrated nitric acid the equation is different. So 3 moles of copper produce 2 moles of NO. Therefore it requires 6 moles of copper to produce 4 moles of NO.
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
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.
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 =========================
This is the Sabatier process and is usually in need of high temperature and a metallic catalyst. So, balanced equation. CO2 + 4H2 -> CH4 + 2H2O As said, hydrogen is in excess, so CO2 is limiting and drives the reaction. One to one 36.6 moles CO2 (1 mole CH4/1 mole CO2) = 36.6 moles CH4 produced in this reaction -------------------------------------------------------
1 mole of H2 reacts with 1 mole of Cl2 to produce 2 moles of HCl. The molar mass of Cl2 is 70.9 g/mol. Since 0.65 g of H2 is used, we must calculate the molar amount of chlorine needed, which is 70.9 g/mol x 0.65 g / 2 g. This gives us approximately 23 grams of chlorine required.
The answer is 3,375 moles oxygen.
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
To calculate the moles of hydrogen needed to produce 68 grams of ammonia (NH₃), we start with the balanced chemical equation for the synthesis of ammonia: N₂ + 3H₂ → 2NH₃. The molar mass of ammonia is approximately 17 g/mol, so 68 grams of NH₃ corresponds to 68 g / 17 g/mol = 4 moles of NH₃. Since 3 moles of hydrogen are required for every 2 moles of ammonia, the moles of hydrogen needed is (4 moles NH₃) × (3 moles H₂ / 2 moles NH₃) = 6 moles of H₂. Therefore, 6 moles of hydrogen must react to produce 68 grams of ammonia.
To find the grams of nitrogen dioxide needed, first calculate the moles of nitrogen monoxide using Avogadro's number. Then, use the balanced chemical equation to determine the moles of nitrogen dioxide required. Finally, convert moles to grams using the molar mass of nitrogen dioxide.
We need 3 moles of potassium perchlorate.
Four moles of potassium chlorate are needed.
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.
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
2KClO3==>2KCl+3O2 is the equation. so you need 4 moles of KClO3.
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).
1 mole of sulfur reacts with 2 moles of oxygen to produce 1 mole of SO2. Therefore, to produce 0.567 moles of SO2, you would need to burn 0.2835 moles of sulfur.
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 =========================