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The above reaction is a combination type reaction between silver (Ag) and sulfur (S)
2 Ag + S = Ag2S
ag is + s is -2
the sliver symbol in the periodic table AG
215.8 g Ag x 1 mol Ag/196.9 = 1.096 moles Ag32.1 g S x 1 mol S/32 g = 1 moles Since Ag and S combine in a ratio of 2:1 (2 Ag to 1 S), the most Ag2S you can form is limited by moles of Ag. 1.096 moles Ag will react with 0.548 moles S ==> Ag2S. So the rest of the S will be left over and not used because there isn't enough Ag. Thus, the mass of resultant salt would be 197 g + 17.5 g = 214.5 g.
16 Ag + S8 -------> 8 Ag2S
The above reaction is a combination type reaction between silver (Ag) and sulfur (S)
The way to solve these problems is to determine which reactant will produce the fewest moles of product. In order to do this, we first need to convert the mass of each reactant into moles of reactant using the molecular mass of silver, Ag (107.9 g/mol), and molecular sulfur, S8 (256.5) g/mol:2.0 g Ag * 1mol/107.9 g/mol = 0.0185 mol Ag2.0 g S8 * 1mol/256.5 g/mol = 0.00779 mol S8Now using the balanced chemical equation (our recipe) from above:16 Ag + S8 -->8 Ag2SWe can see:16 moles of Ag turns into 8 moles of Ag2S1 mole of S8 turns into 8 moles of Ag2SWhat we need to do know, is use these two statements above to see how many moles of product each reactant turns into:0.0185 mol Ag * 8 mol Ag2S/16 mol Ag = 0.00925 mol Ag2S0.00779 mol S8 * 8mol Ag2S/1 mol S8 = 0.0623 mol Ag2SSince 0.0185 mole of Ag produce less Ag2S (the product), silver (Ag) is the limiting reactant. Now you could convert this to grams of product by multiplying the lesser amount produced, 0.00925 moles of Ag2S, by the molecular mass of Ag2S.To determine the mass of reactant left unreacted, we need to convert moles of the limiting reactant into moles of the other reactant to find out how much of the other reactant is lost--Then subtract from the original amount and convert to grams with the molecular mass.0.0185 mol Ag * 1 mol S8/16 mol Ag = 0.00116 mol S8Originally we had 0.00779 mol of S8, from above, so:0.00779 mol S8 - 0.00116 mol S8 = 0.00663 mol S8Now convert to grams left:0.00663 mol S8 * 256.5 g/mol = 1.70 g S8 left unreacted!
1 mol Ag/ 107.87g Ag ---/---------------------------------------- x2=215.74g Ag / 1mol Ag 1 mol S/32.07g S --------/------------------------------------32.07g S /1 mol S total=247.81g Ag2S 215.47g Ag/247.81=.8706 87.06% Ag .8706 or 87.06% Ag x 125g Ag2S = 108.83g Ag can be produced from Ag2S
The balanced equation is as follows: 2Ag + H2S --> Ag2S + H2
2 Ag + S = Ag2S
ag is + s is -2
the sliver symbol in the periodic table AG
215.8 g Ag x 1 mol Ag/196.9 = 1.096 moles Ag32.1 g S x 1 mol S/32 g = 1 moles Since Ag and S combine in a ratio of 2:1 (2 Ag to 1 S), the most Ag2S you can form is limited by moles of Ag. 1.096 moles Ag will react with 0.548 moles S ==> Ag2S. So the rest of the S will be left over and not used because there isn't enough Ag. Thus, the mass of resultant salt would be 197 g + 17.5 g = 214.5 g.
yes it is Ag plus S (Sulfur found in the atmosphere) yeilds Ag2S silver sulfide
Ag2S is Ionic. If it has a Non-Metal and a Metal it is Ionic. If it only contains non-metals it is noniconic. Well that simple rule is OK- however a better guess is made if you conside the electronegativities of Ag and S - they are quite close (they need to be well apart for ionic bonding) so Ag2S has considerable covalent character.
The molecular formula for Silver sulfide is Ag2S