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What mass of barium sulfate can be produced when 100.00 ml of a 100 M solution of barium chloride is mixed with 100.0 ml of a 100 M solution of iron III sulfate?
To find the mass of barium sulfate produced, first calculate the moles of barium chloride and iron III sulfate using their volumes and concentrations. Then, determine the limiting reactant by comparing the moles of barium chloride and iron III sulfate and use it to find the moles of barium sulfate produced. Finally, multiply the moles of barium sulfate by its molar mass to find the mass in grams.
How many grams of silver chloride are formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride?
To find the limiting reactant, we need to determine how many grams of silver chloride can be produced from each reactant and compare the results. Calculate the amount of silver chloride that can be produced from 10.0 g of silver nitrate. Calculate the amount of silver chloride that can be produced from 15.0 g of barium chloride. The reactant that produces the lesser amount of silver chloride will be the limiting reactant.
With 188 garms of barium chloride there would be how much 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.
How many grams of solid barium sulfate form when 22.6 mL of 0.160 M barium chloride reacts with 54.6 mL of 0.055 M sodium sulfate?
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
How many grams of chlorine gas are needed to make 117 grams of sodium chloride?
117 grams of sodium chloride (NaCl) is equivalent to 117 grams of chlorine gas because each molecule of NaCl contains one sodium atom and one chlorine atom.
How many grams of barium chloride are needed to make 100. grams of barium sulfate?
89.3
What mass of barium sulfate can be produced when 100.00 ml of a 100 M solution of barium chloride is mixed with 100.0 ml of a 100 M solution of iron III sulfate?
To find the mass of barium sulfate produced, first calculate the moles of barium chloride and iron III sulfate using their volumes and concentrations. Then, determine the limiting reactant by comparing the moles of barium chloride and iron III sulfate and use it to find the moles of barium sulfate produced. Finally, multiply the moles of barium sulfate by its molar mass to find the mass in grams.
How many grams of barium sulfate can be produced from 20.8 grams of barium sulfate?
If you start with 20.8 grams of barium sulfate (BaSO₄), you cannot produce more barium sulfate from it; you can only measure how much you have. Therefore, you can produce a maximum of 20.8 grams of barium sulfate if you are referring to using the same amount of BaSO₄ in a reaction or process. In summary, you have 20.8 grams of barium sulfate available, not more.
How many grams of HCl are needed to make 21.5 grams of Na2SO4?
Sodium sulfate is not prepared from hydrogen chloride.
When silver nitrate reacts with barium chloride silver chloride and barium nitrate are formed How many grams of silver chloride are formed when 10.8 g of silver nitrate reacts with 15 g of barium?
9.11 g
How many grams of barium chloride are there in 286 moles?
To convert moles to grams, you need to use the molar mass of barium chloride, which is 208.23 g/mol. Multiply the number of moles (286 moles) by the molar mass to find the grams of barium chloride: 286 moles * 208.23 g/mol = 59,545.78 grams.
How many liters of water are required to dissolve 1 of barium sulfate?
Barium sulfate (BaSO₄) is known for its very low solubility in water, with a solubility product (Ksp) indicating that only about 0.0002 grams can dissolve in 1 liter of water at room temperature. Therefore, to dissolve 1 gram of barium sulfate, an impractically large volume of water—approximately 5,000 liters—would be needed. This highlights the compound's nature as a sparingly soluble salt.
How many grams of silver chloride can be produced if you start with 4.62 grams of barium chloride 2AgNO3 plus BaCl2 and mdash and gt 2AgCl plus Ba(NO3)?
6,36 g of silver chloride are obtained.
What is the molar mass of barium sulfate?
233.43 g On A+ the answer's 233.4 =]
How many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate assuming that you have an adequate amount of lead sulfate to do the reaction?
To determine the number of grams of lithium nitrate needed to make 250 grams of lithium sulfate, you need to calculate the molar mass of lithium sulfate and lithium nitrate, then use stoichiometry to find the ratio of lithium nitrate to lithium sulfate. Finally, apply this ratio to find the mass of lithium nitrate needed for the reaction. Lead sulfate is not involved in this calculation as it is not part of the reaction between lithium nitrate and lithium sulfate.
How many grams of lithium nirate will be needed to make 250 grams of lithium sulfate assuming that you have an adequate amount of lead sulfate to do the reaction?
To find out the grams of lithium nitrate needed, you need to calculate the molar mass of lithium sulfate (Li2SO4) and lithium nitrate (LiNO3). Then use stoichiometry to determine the amount of lithium nitrate required to produce 250 grams of lithium sulfate. The balanced chemical equation for the reaction would also be needed.
How many grams of silver chloride are formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride?
To find the limiting reactant, we need to determine how many grams of silver chloride can be produced from each reactant and compare the results. Calculate the amount of silver chloride that can be produced from 10.0 g of silver nitrate. Calculate the amount of silver chloride that can be produced from 15.0 g of barium chloride. The reactant that produces the lesser amount of silver chloride will be the limiting reactant.
