determine the mass of water formed when 12.5 l nh3,

In this article, we will be looking at the amount of water that can form when 12.5 liters of ammonia gas is reacted with oxygen gas to create aqueous ammonia and water. This reaction is called water-gas shift. It occurs in the gas phase, and it proceeds to make ammonia (NH_) from hydrogen (H_), nitrogen (N_), and oxygen (O_). The equation for this process looks like: \begin{align*} 12.500\; L ~~ NH_{}^{+}\leftrightarrow HONH_{}^{+ OH^- O^- + -}{g}}{\end{align*}. When these reactions occur we can see that there are three products of them, aqueous ammonia (.OH ), Hydrogen oxide ({HO}) also known as water vapor or steam, and Oxygen atoms (). This means that when enough of each reactant is present, aqueous ammonia will form. In order to determine how much water is formed in the reaction, we can use mass balance equation: \begin{align*} 12.500 ~~ NH_{}^{+}\leftrightarrow HONH_{}^{+ OH^- O^- + -}{g}}{\end{align*}. We know that one liter of pure aqueous ammonia has an equal weight of 0.0625 kg (or 625 grams). This means that when there are enough reactants for this process to occur, the total amount of water produced per gram of hydrogen and nitrogen gas would be about 2700 grams or 7650 milliliters (.76 L) or nearly 15 lit