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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.
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How many grams of barium chloride are needed to make 100. grams of barium sulfate?
89.3
How many grams of barium sulfate can be produced from 20.8 g of barium chloride?
To determine the grams of barium sulfate produced from 20.8 g of barium chloride, we first need to consider the balanced chemical reaction between barium chloride (BaCl₂) and sodium sulfate (Na₂SO₄) to produce barium sulfate (BaSO₄) and sodium chloride (NaCl). The molar mass of barium chloride is approximately 137.33 g/mol (for Ba) + 2 × 35.45 g/mol (for Cl) = 137.33 + 70.90 = 208.23 g/mol. The molar mass of barium sulfate is about 233.39 g/mol. Using stoichiometry: Calculate moles of BaCl₂: ( \text{moles} = \frac{20.8 , \text{g}}{208.23 , \text{g/mol}} \approx 0.100 , \text{mol} ). Since the reaction produces 1 mole of BaSO₄ for every mole of BaCl₂, 0.100 moles of BaCl₂ will produce 0.100 moles of BaSO₄. Convert moles of BaSO₄ to grams: ( 0.100 , \text{mol} \times 233.39 , \text{g/mol} \approx 23.34 , \text{g} ). Thus, approximately 23.34 grams of barium sulfate can be produced from 20.8 grams of barium chloride.
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.
How many moles are 4.12 grams aluminum sulfate?
4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous salt).
How many grams of barium chloride are needed to make 100 grams of barium sulfate?
Since barium sulfate and barium chloride have a 1:1 molar ratio, you would need the same amount of barium chloride as barium sulfate, so 100 grams.
How many grams of barium chloride are needed to make 100. grams of barium sulfate?
89.3
How many grams of barium sulfate can be produced from 20.8 g of barium chloride?
To determine the grams of barium sulfate produced from 20.8 g of barium chloride, we first need to consider the balanced chemical reaction between barium chloride (BaCl₂) and sodium sulfate (Na₂SO₄) to produce barium sulfate (BaSO₄) and sodium chloride (NaCl). The molar mass of barium chloride is approximately 137.33 g/mol (for Ba) + 2 × 35.45 g/mol (for Cl) = 137.33 + 70.90 = 208.23 g/mol. The molar mass of barium sulfate is about 233.39 g/mol. Using stoichiometry: Calculate moles of BaCl₂: ( \text{moles} = \frac{20.8 , \text{g}}{208.23 , \text{g/mol}} \approx 0.100 , \text{mol} ). Since the reaction produces 1 mole of BaSO₄ for every mole of BaCl₂, 0.100 moles of BaCl₂ will produce 0.100 moles of BaSO₄. Convert moles of BaSO₄ to grams: ( 0.100 , \text{mol} \times 233.39 , \text{g/mol} \approx 23.34 , \text{g} ). Thus, approximately 23.34 grams of barium sulfate can be produced from 20.8 grams of barium chloride.
How many moles of barium sulfate are produced from 0.100 mole of barium chloride?
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.
How many grams of aluminum sulfate are produced from 2.67 mol sulfuric acid?
To find the grams of aluminum sulfate produced, you would need to know the stoichiometry of the reaction between sulfuric acid and aluminum sulfate. Without that information, we cannot determine the exact amount of aluminum sulfate produced.
What happens when barium sulfate is added with water?
When barium sulfate is added to water, it forms a suspension in which the particles do not dissolve in the water. This suspension is not soluble in water and can be filtered out. Barium sulfate is almost insoluble in water, which makes it useful for certain medical tests such as barium sulfate contrast studies.
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 elements form a molecule of barium sulfate?
A molecule of barium sulfate consists of one barium atom (Ba) and one sulfur atom (S), combined with four oxygen atoms (O), for a total of six elements.
How many moles are 4.12 grams aluminum sulfate?
4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous 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.
How many grams are in one mole of sulfate?
The molar mass of sulfate is 96.06 g/mol. Therefore, one mole of sulfate contains 96.06 grams of sulfate.
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
