and he just started this lesson and Dunt understand anyting on this type of stiochiometry. would be greatly appreciated if sum1 gave me an explanation / instructions for each stage of manufacture rather than simply show all the work.
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First, write the balanced equation. This is very importante.2AgNO3 Ag2S + K2S ==> + 2KNO3Ahora all reactions are based on the moon rather than grams, so 2 moles of AgNO3 reacts with one mole K2S to produce one mole of Ag2S and 2 moles KNO3. Grams that are given in the problem, you can convert everything to moles grams and convert it back to the end, because the question asks gramos.Masa molar silver nitrate = 169. 9 g / sulfide molar potassium molMasa = 110. 3 g / mole silver sulfide molMasa = 247. 8 g / gm mol15 AgNO3/169. 9 g / mol = 0. 088 moles GM K2S/110 AgNO315. 3 g / mol = 0. 136 moles of K2SPare here for a minute and notice that there is a smaller number of moles of AgNO3 as K2S. Back to the equation. Therefore, AgNO3 is called the limiting reagent, because escasez.También must not forget that you need two moles of AgNO3 to a mol Ag2S to 0. 088 moles of AgNO3 is available only 0. 044 moles of Ag2S. Now we must make this 0. 044 moles of Ag2S in gramos.247. 8 g / mol x 0. 044 moles = 10. 9 grams of Ag2S occur.
First, you must write the balanced chemical equation. It is important to be balanced, because we will use mole ratios to solve this problem: 2AgNO3 (aq) + K2S (aq) -> Ag2S (s) 2KNO3 + (aq) Therefore, by examining the coefficients balanced equation, their molar ratios, respectively, are 2, 1, 1, 2. We will refer to these more useful tarde.Su record all the information available to it, if you type gives the mass (m) of respectivo.m AgNO3m = 15g = 15 g K2STambién can fill molar mass (M), which can be easily discovered by looking at your table periódica.M AgNO3 = 107. 87 g / mol 14 +. 00 g / mol + (16. 00 g / mol * 3) = 169. 87 g / MolMed of K2S = (39. 10 g / mol * 2) + 32. 06 g / mol = 110. 26 g / molAhora enough information to determine the number of moles (n) for both, according to the formula n = m / mn = 15g/169 AgNO3. 87 g / mol = 0. 08830282 Molné of K2S = 15g/110. 26 g / mol = 0. 1304208 Molné round, however, we are still in the data processing bruto.Así now that we have the number of moles of silver nitrate and potassium sulfide, one can now calculate the limiting reagent using relations molares . Así that first gives a relationship between potassium and sulfur and silver nitrate, which is 1:2. (Whatever you choose, you can make silver nitrate and potassium sulfide in a 2:1 ratio.) Use of moles Potassium sulfide is earlier: K2S n = 0. 1304208 molPodemos use mole ratios to find the limiting reagent: 0. K2S 1304208 2 mol AgNO3 * mol / 1 mol K2SEl objective is to obtain the potassium sulfide to replace, in this case, it does. This is where it gets a little confused. Since the number of moles (0. … 27 208 AgNO3 Mol) is more than the number of moles, calculated above, is the silver nitrate reagent limitante.Por So now we can establish another relationship between molar silver nitrate and silver sulfide, which is 2:1. This means that we must find the moles with the molar ratios again. Thus, by the same method as above and you should get 0. 04415141 mol. Now we have the number of moles of silver sulfide and molecular weight (which can easily be calculated.) Then rearrange the formula M n = m / MB masa.m clear = = NMM (0. 04,415,141 mol) (247. 80 g / mol) m = 10. 94 g, and then return the correct number of significant digits, which is equal to the lesser number of significant figures in a given position on an issue. In this case, because 15g is 2 to 2 digits significativos.m g.Por = 11 Therefore, the mass of silver sulfide is 11 g.Revisa alone, I could be wrong, but the process of how you ask these questions.