What is the mass (in grams) of water that is produced when 100.0 grams of nitrogen monoxide is reacted with 75.0 grams of hydrogen according tho the following balanced eguation?
2NO + 5H2 = 2NH3 + 2H2O
As reactants, you have 100/30 = 3.33 moles of NO and 75/2 = 37.5 moles of H2. Clearly, NO is the limiting reactant and there is excess H2. Only 3.33 moles of H2O can be formed. Convert that to grams.
To determine the mass of water produced, we need to use the stoichiometry of the balanced equation.
First, calculate the number of moles of each substance using their respective molar masses.
The molar mass of nitrogen monoxide (NO) is 30.0 g/mol.
The molar mass of hydrogen (H2) is 2.02 g/mol.
The molar mass of water (H2O) is 18.02 g/mol.
Moles of nitrogen monoxide (NO) = mass (in grams) / molar mass
Moles of nitrogen monoxide (NO) = 100.0 g / 30.0 g/mol = 3.33 mol
Moles of hydrogen (H2) = mass (in grams) / molar mass
Moles of hydrogen (H2) = 75.0 g / 2.02 g/mol = 37.13 mol
Based on the balanced equation, the mole ratio between nitrogen monoxide (NO) and water (H2O) is 2:2. This means that for every 2 moles of NO, 2 moles of water are produced.
To find the moles of water produced, we use the mole ratio and the moles of nitrogen monoxide as the given value:
Moles of water (H2O) = moles of nitrogen monoxide (NO) * (moles of water / moles of nitrogen monoxide) = 3.33 mol * (2 mol H2O/2 mol NO) = 3.33 mol
Finally, calculate the mass of water produced using the moles of water and the molar mass of water:
Mass of water (H2O) = moles of water (H2O) * molar mass of water
Mass of water (H2O) = 3.33 mol * 18.02 g/mol = 59.97 g
Therefore, the mass of water produced when 100.0 grams of nitrogen monoxide reacts with 75.0 grams of hydrogen is approximately 59.97 grams.