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.
4.67
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
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 -------------------------------------------------------
To calculate the number of moles from grams, you must divide by the substance's molar mass
4.67
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
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
We need 3 moles of potassium perchlorate.
Four moles of potassium chlorate are needed.
2KClO3==>2KCl+3O2 is the equation. so you need 4 moles of KClO3.
How many grams of nitrogen dioxide must react with water to produce 5.00 x 1022 molecules of nitrogen monoxide?
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 =========================
2CsClO4 -> 2Cs + Cl2 + 4O2 2.7 moles O2 (2 moles CsClO4/4 mole O2) = 1.4 mole CsClO4 needed
Given the balanced equation Kr + 3F2 --> KrF6 In order to find how many moles of F2 are needed to produce 3.0 moles of KrF6, we must convert from moles to moles (mol --> mol conversion). 3.0 mol KrF6 * 3 molecules F2 = 9.0 mol F2 --------- 1 molecule F2
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