determine the amount of water formed in 4NH3 + 5O2 4NO + 6H2O if 10 g of Oxygen is mixed with 10g nitrigen trihydride in grams
To determine the amount of water formed in the reaction 4NH3 + 5O2 -> 4NO + 6H2O, we need to follow a few steps. Here's how you can do it:
1. Convert the given masses of Oxygen (O2) and Nitrogen Trihydride (NH3) to moles. To do this, divide the given masses by their respective molar masses. The molar mass of O2 is 32 g/mol, and the molar mass of NH3 is 17 g/mol.
The moles of Oxygen (O2) = 10 g / 32 g/mol = 0.3125 mol
The moles of Nitrogen Trihydride (NH3) = 10 g / 17 g/mol = 0.5882 mol
2. Determine the limiting reagent. The limiting reagent is the reactant that will be completely consumed and controls the amount of product formed. To find the limiting reagent, compare the moles of each reactant to the stoichiometric ratio given in the balanced equation. The stoichiometric ratio in the balanced equation is 4:5 for NH3 to O2.
Using the stoichiometric ratio, we can calculate the number of moles of NH3 needed for the reaction: 0.3125 mol O2 * (4 mol NH3 / 5 mol O2) = 0.25 mol NH3.
Since we have 0.5882 mol of NH3, which is greater than 0.25 mol needed, NH3 is in excess. Therefore, Oxygen is the limiting reagent.
3. Determine the moles of water formed. According to the balanced equation, for every 5 moles of Oxygen (O2), 6 moles of water (H2O) are produced.
Using the stoichiometric ratio, we can calculate the number of moles of water formed: 0.3125 mol O2 * (6 mol H2O / 5 mol O2) = 0.375 mol H2O.
4. Convert the moles of water to grams. To do this, multiply the moles of water by its molar mass. The molar mass of water (H2O) is 18 g/mol.
The grams of water formed = 0.375 mol H2O * 18 g/mol = 6.75 g.
Therefore, the amount of water formed when 10 g of Oxygen is mixed with 10 g of Nitrogen Trihydride is 6.75 grams.