When 36.0 grams of water, H2O, react, the mole ratio of nitrogen (N2) and hydrogen (H2) produces two moles of ammonia (NH3). This means that for every 4 NH3 molecules produced, 6 H2O molecules are consumed. Since the molar mass of H2O is 18 g/mol, this means that for every 36.0 g of H2O, 18.0 g of NH3 are produced.

To calculate the exact number of grams of NH3 produced in this reaction, we can use the mole ratio of H2 to NH3. According to the third web search result, the mole ratio of hydrogen to ammonia is 3 H : 1 NH3. To calculate the number of grams of NH3 produced, we need to multiply the number of moles of H2 (which is 2) by the mole ratio of hydrogen to ammonia (3 H : 1 NH3), and then multiply the product by the molar mass of ammonia (17 g/mol). This gives us a total of 102 g of NH3 produced from 36.0 g of H2O.

In summary, when 36.0 g of H2O react, 102 g of NH3 are produced. This is due to the mole ratio of nitrogen (N2) and hydrogen (H2) producing two moles of ammonia (NH3), and the mole ratio of H2 to NH3 being 3 H : 1 NH3. The molar mass of H2O (18 g/mol) and NH3 (17 g/mol) further help to calculate the exact number of grams of NH3 produced.