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03:18 University of Maine
Problem 116

# 3.116 For the reaction between solid tetraphosphorus trisulfide and oxygen gas to form solid tetraphosphorus decoxide and sulfur dioxide gas, write a balanced equation. Show the equation (seeTable 3.4 ) in terms of (a) molecules, (b) moles, and (c) grams.

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## Video Transcript

give it a description of a chemical reaction. This could be written as a balanced chemical equation, which can then be interpreted in several different ways. So if, for example, we have solid tetro, phosphorus try sulfide reacts with oxygen gas to form solid tetra phosphorus dick oxide and sulphur dioxide gas. Our first step is to write the chemical formulas for each of these substances. Tetra phosphorus try sulfide contains phosphorus and tetra means four, and sulfide is so for, and there are three because of the try two solid plus oxygen gas in the chemical reaction. Oxygen gas is always +02 and it forms solid tetra phosphorus. So again, P four Decca oxide deca is the prefix for 10 plus sulphur dioxide, which is one sulfur and to Oxygen's. This is a gas, and this is a solid. Once we've written our skeletal equation, we can balance the equation by changing the coefficients we see on this side. Therefore, phosphorus is on this side. There for phosphorus is here. There are three suffers, and here there's only one, so we multiply by three. This gives us 10 oxygen's plus three more here for a total of 16 So that means the coefficient here has to be eight. So now we have a balanced equation, and we can interpret this in several different ways. If we think of it. In terms of molecules, this equation tells us the ratio of molecules that will form. So, for example, if we have one molecule of P for s three, it will react with eight molecules of oxygen to form one molecule of P for 0 10 and three molecules of sulfur dioxide. So the coefficients from the balanced equation can tell us how many molecules react with each typically because we don't do reactions and molecules, they're too small to measure out. We use moles instead. On this equation can also describe the number of moles required. So, for example, the coefficients tell us that one mole of P for S three we'll react with eight moules of 02 to form one more p for 0 10 and one mole of sulfur dioxide so it can be interpreted in terms of other molecules or moles. The third way that we can consider it is to look at it in terms of mass in the law of conservation of mass as well as the relationship between mass and moles. In order to do this, we need to know the molar mass of each substance. And so we find that by using the periodic table So one more of P four s three has the molar mass before times the molar mass of phosphorus, plus three times the molar mass of sulfur, or 200 and 20 point 093 grams. Similarly, we confined the molar mass of oxygen to be 31 0.998 grams by adding two times 15 and the molar mass of P for 0 10 is equal to 200 and 83 0.886 grams. And finally, the molar mass of sulfur dioxide is equal to 64 0.58 grams. And so we can interpret this equation in terms of grams by multiplying the molar mass by the coefficient in front of the equation. So we multiply this number by eight in this number by three, and these still are multiplied by one. So for the reaction, I need to have this many grams of each substance, and I could describe the reaction in terms of grams so it would be 220 grams of p for S three reacts with 255 went 984 grams of oxygen to form 283 grams of P for 0 10 and 192 0.174 grams of sulfur dioxide.