Since both barium chloride and barium sulfate contain one mole of barium atoms pert mole of compound, the moles of barium sulfate will be the same, 0.100, when barium has the limiting concentration in the production of the sulfate.
The formula of barium nitrate is Ba(NO3)2 . . . Therefore, there are 0.01 moles of Barium in Barium Nitrate
1
1.00 mol
I assume you mean 0.5 molar, but I'll go with what I'm given. If you have 500 molar of barium chloride, you will have 1000 molar of chloride ions as there are 2 chlorides for every 1 molecule of barium chloride. Having the 100ml there is irrelevant as you are talking about concentration and didn't ask for moles.
Balanced equation first. BaCl2 + Na2SO4 -> 2NaCl + BaSO4 22.6 ml BaCl2 = 0.0226 liters 54.6 ml Na2SO4 = 0.0546 liters 0.160 M BaCl2 = moles BaCl2/0.0226 liters = 0.00362 moles BaCl2 0.055 M Na2SO4 = moles Na2SO4/0.0546 liters = 0.0030 moles Na2SO4 The ratio of BaCl2 to Na2SO4 is one to one, so either mole count wull drive this reaction. Use 0.0003 moles Na2SO4 0.0030 moles Na2SO4 (1 mole BaSO4/1 mole Na2SO4)(233.37 grams/1 mole BaSO4) = 0.700 grams of BaCO4 produced
Since both chloride anions and nitrate anions have a charge of -1, there will be the same number of moles of silver chloride produced as the moles of silver nitrate reacted. (Since both silver nitrate and silver chloride are ionic compounds, it would be preferable to call their "moles" "formula units" instead.)
4 moles
First, you want to figure out how many grams 1 mol of barium is. Look at the periodic table of elements. (www.webelements.com) Barium is Ba. Scroll down a little bit after clicking and you will see atomic weight. It is 137g. Now get the atomic weight for chlorine. It is Cl. The atomic weight is 35.5g. Just add them together. 137+35.5 = 172.5g = 1 mol of barium chloride. For 188g, just divide it by 172.5g. Which is 1.09 moles.
3,75 moles barium chloride
1020g
The molar mass of barium sulfate, BaSO4 is 233.4 Amount of BaSO4 = mass of sample / molar mass = 14.2/233.4 = 0.0608mol 0.0608 moles of BaSO4 are conained in a 14.2g pure sample.
The Ksp for BaS04, which is barium sulfate, is 1.1 x 10^-10. Ksp is the solubility product. It is the product of the solubility of the ions in moles per liter.
1,075 moles of aluminium chloride are obtained.
I assume you mean 0.5 molar, but I'll go with what I'm given. If you have 500 molar of barium chloride, you will have 1000 molar of chloride ions as there are 2 chlorides for every 1 molecule of barium chloride. Having the 100ml there is irrelevant as you are talking about concentration and didn't ask for moles.
Balanced equation first. BaCl2 + Na2SO4 -> 2NaCl + BaSO4 22.6 ml BaCl2 = 0.0226 liters 54.6 ml Na2SO4 = 0.0546 liters 0.160 M BaCl2 = moles BaCl2/0.0226 liters = 0.00362 moles BaCl2 0.055 M Na2SO4 = moles Na2SO4/0.0546 liters = 0.0030 moles Na2SO4 The ratio of BaCl2 to Na2SO4 is one to one, so either mole count wull drive this reaction. Use 0.0003 moles Na2SO4 0.0030 moles Na2SO4 (1 mole BaSO4/1 mole Na2SO4)(233.37 grams/1 mole BaSO4) = 0.700 grams of BaCO4 produced
Since both chloride anions and nitrate anions have a charge of -1, there will be the same number of moles of silver chloride produced as the moles of silver nitrate reacted. (Since both silver nitrate and silver chloride are ionic compounds, it would be preferable to call their "moles" "formula units" instead.)
4 moles
AgNO3 + NaCl ===> AgCl(s) + NaNO37 moles silver nitrate will produce 7 moles of silver chloride provided there is sufficient (at least 7 moles) of sodium chloride.
First, you want to figure out how many grams 1 mol of barium is. Look at the periodic table of elements. (www.webelements.com) Barium is Ba. Scroll down a little bit after clicking and you will see atomic weight. It is 137g. Now get the atomic weight for chlorine. It is Cl. The atomic weight is 35.5g. Just add them together. 137+35.5 = 172.5g = 1 mol of barium chloride. For 188g, just divide it by 172.5g. Which is 1.09 moles.
If the reaction were 2AlCl3 + 3Pb ---> 3PbCl2 + 2Al, then from this balanced equation, 14 moles of AlCl3 would produce 3/2 x 14 moles of PbCl2 = 21 moles of PbCl2.