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Topic: **Chemistry homework help balancing equations****Question:**
1. Silver nitrate reacts with calcium chloride to produce silver chloride and calcium nitrate. What mass of silver chloride can be formed from 47.7g of calcium chloride?
2. A reaction mixture contains 11.0 g of PCl3 and 6.83 g of PbF2.
PbF2 + PCl3 → PF3 + PbCl2
How many grams of the compound in excess remains after complete reaction?

June 16, 2019 / By Laureen

1. The amount of chloride in 47.7 g of calcium chloride is *exactly* the same as the amount of chloride in X g of silver chloride. Just come up with the ratio of molar weights per atom of chlorine, and multiply it by 47.7 2. You need to balance the equation first 3 PbF2 + 2 PCl3 => 2 PF3 + 3 PbCl2 So figure out how much 3 moles of PbF2 weights, and how much 2 moles of PCl3 weighs. Compare that ratio to the actual mass you have. You will have x% of 3 moles of PbF2 and (x+y)% of 2 moles of PCl3; y, however, may be positive or negative. If y is positive, then you have a surplus of PCl3, and y% of 2 moles of PCl3 is your answer. If y is negative, then you have a surplus of PbF2, and -y% of 3 moles of PbF2 is your answer.

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1. The amount of chloride in 47.7 g of calcium chloride is *exactly* the same as the amount of chloride in X g of silver chloride. Just come up with the ratio of molar weights per atom of chlorine, and multiply it by 47.7 2. You need to balance the equation first 3 PbF2 + 2 PCl3 => 2 PF3 + 3 PbCl2 So figure out how much 3 moles of PbF2 weights, and how much 2 moles of PCl3 weighs. Compare that ratio to the actual mass you have. You will have x% of 3 moles of PbF2 and (x+y)% of 2 moles of PCl3; y, however, may be positive or negative. If y is positive, then you have a surplus of PCl3, and y% of 2 moles of PCl3 is your answer. If y is negative, then you have a surplus of PbF2, and -y% of 3 moles of PbF2 is your answer.

1. First of all, you need to write out your equation to find out the mole ratio of CaCl to AgCl in this reaction. We find that the equation is AgNO3 + CaCl = AgCl +CaNO3. This means that the mole ratio of CaCl to AgCl is 1:1. This makes it much easier for us. Next you have to convert 47.7g of CaCl to moles by dividing the number of grams by the molar mass. CaCl has a molar mass of 75.533g/mol, so we have about 0.6315 moles of CaCl. Since the mole ratio is 1:1, we will also have 0.6315 moles of AgCl. Now we have to convert back from moles to grams by multiplying the molar mass by the number of moles. AgCl has a molar mass of 143.321 g/mol, so your final answer is 90.5 g (3 significant figures) of AgCl. 2. The balanced equation here is 3PbF2 +2PCl3 = 2PF3 + 3PbCl2. Here again you convert both from grams to moles. You get 0.0279 moles of PbF2 and 0.0801 moles of PCl3. Now, looking back at the balanced equation, we need 3 moles of PbF2 for every 2 moles of PCl3, or 1.5 times as many moles of PbF2. 0.0279 is obviously less than 0.0801, so PbF2 will be our limiting reactant. All of the PbF2 will react, and 2/3 that much PCl3 will react. Now the math is easy. 2/3 of .0279 is 0.0186. This means that only 0.0186 moles of PCl3 will react, leaving you with 0.0615 moles of PCl3. Converting that to grams leaves you with 8.45 grams of compound left in excess.

1. First of all, you need to write out your equation to find out the mole ratio of CaCl to AgCl in this reaction. We find that the equation is AgNO3 + CaCl = AgCl +CaNO3. This means that the mole ratio of CaCl to AgCl is 1:1. This makes it much easier for us. Next you have to convert 47.7g of CaCl to moles by dividing the number of grams by the molar mass. CaCl has a molar mass of 75.533g/mol, so we have about 0.6315 moles of CaCl. Since the mole ratio is 1:1, we will also have 0.6315 moles of AgCl. Now we have to convert back from moles to grams by multiplying the molar mass by the number of moles. AgCl has a molar mass of 143.321 g/mol, so your final answer is 90.5 g (3 significant figures) of AgCl. 2. The balanced equation here is 3PbF2 +2PCl3 = 2PF3 + 3PbCl2. Here again you convert both from grams to moles. You get 0.0279 moles of PbF2 and 0.0801 moles of PCl3. Now, looking back at the balanced equation, we need 3 moles of PbF2 for every 2 moles of PCl3, or 1.5 times as many moles of PbF2. 0.0279 is obviously less than 0.0801, so PbF2 will be our limiting reactant. All of the PbF2 will react, and 2/3 that much PCl3 will react. Now the math is easy. 2/3 of .0279 is 0.0186. This means that only 0.0186 moles of PCl3 will react, leaving you with 0.0615 moles of PCl3. Converting that to grams leaves you with 8.45 grams of compound left in excess.

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The valence "ring" or maximum outer ring is what bonds in the bonding of atoms. ONE considerable rule is that the outer ring would desire to have 8 electrons to be bonded. SO shall we are saying we are wanting to style a bond between fluoride (has 7 valence electrons) and Hydrogen (a million valence electron) they'd create a ideal bond. if we would have enjoyed to bond hydrogen and oxygen(6 valence) at the same time we would desire 2 hydrogen for a million oxygen. in actuality take the 8 entire you desire and subtract between the chemical ingredients faraway from it case in point with the oxygen and hydrogen>> 8 entire mandatory-6 in oxygen= 2 mandatory .

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read your textbook do your own work. i remember when i was in highschool some kids managed to pass thier classes on their own doing that. and the kids that cheated guess what they ended up doing? working at mcdonalds while everyone else went off to college.

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