Final answer:
By using stoichiometry and the balanced chemical equation for the reaction of ammonia and oxygen, we calculate that the reaction of 4.1 g of ammonia will produce 6.504 grams of water.
Step-by-step explanation:
Stoichiometric Calculation for Water Production
To determine the mass of water produced from the reaction of ammonia with oxygen, we need the balanced chemical equation for the reaction and the molar mass of ammonia (NH3) and water (H2O). The chemical reaction is as follows:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
Now let's calculate the moles of ammonia. We know that the molar mass of ammonia (NH3) is 17.031 g/mol. Therefore,
moles of NH3 = mass of NH3 / molar mass of NH3 = 4.1 g / 17.031 g/mol = 0.2407 moles of NH3
According to the balanced equation, 4 moles of NH3 produce 6 moles of H2O. Using stoichiometry, we calculate the moles of H2O produced:
moles of H2O = (0.2407 moles NH3) x (6 moles H2O / 4 moles NH3) = 0.3611 moles H2O
Next, we use the molar mass of water (H2O), which is 18.015 g/mol, to find the mass of water produced:
mass of H2O = moles of H2O x molar mass of H2O = 0.3611 moles x 18.015 g/mol = 6.504 g of H2O
Therefore, the reaction of 4.1 g of ammonia with oxygen gas will produce 6.504 grams of water.