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To find the number of moles of sulfur in 2.55g of aluminum sulfate, you need to first calculate the molar mass of aluminum sulfate (Al2(SO4)3), which is 342.15 g/mol. Sulfur accounts for 3 moles in one mole of aluminum sulfate, so you can calculate the number of moles by dividing the given mass (2.55g) by the molar mass of aluminum sulfate.
To convert grams of aluminum sulfate to moles, you first need to determine the molar mass of aluminum sulfate (Al2(SO4)3), which is approximately 342.15 g/mol. Then, divide the given mass by the molar mass to obtain the number of moles. In this case, 6.7 grams of aluminum sulfate is approximately 0.02 moles.
To determine the limiting reactant, we need to compare the moles of each reactant. First, calculate the moles of aluminum and copper sulfate separately. Then, determine the mole ratio between them and see which reactant is present in lower amount compared to the stoichiometric ratio. The reactant that is present in lower moles is the limiting reactant.
The molar mass of aluminum sulfate is 342.15 g/mol. Therefore, the mass of 0.25 moles of aluminum sulfate would be 85.54 grams (0.25 moles x 342.15 g/mol).
There are 5 moles of sulfur in 5 moles of H2SO4, as there is 1 mole of sulfur in each mole of H2SO4.
To find the number of moles of sulfur in 2.55g of aluminum sulfate, you need to first calculate the molar mass of aluminum sulfate (Al2(SO4)3), which is 342.15 g/mol. Sulfur accounts for 3 moles in one mole of aluminum sulfate, so you can calculate the number of moles by dividing the given mass (2.55g) by the molar mass of aluminum sulfate.
To find the mass of 0.25 moles of aluminum sulfate, you need to know the molar mass of aluminum sulfate. The molar mass of aluminum sulfate (Al2(SO4)3) is approximately 342.15 g/mol. Therefore, the mass of 0.25 moles of aluminum sulfate would be around 85.54 grams.
4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous salt).
To find the moles of sulfate ions in the solution, you first need to determine the moles of aluminum sulfide present. Since you are not provided with the amount of aluminum sulfide, you cannot calculate the moles of sulfate ions. Additionally, oxygen is not relevant to determining the moles of sulfate ions.
To convert grams of aluminum sulfate to moles, you first need to determine the molar mass of aluminum sulfate (Al2(SO4)3), which is approximately 342.15 g/mol. Then, divide the given mass by the molar mass to obtain the number of moles. In this case, 6.7 grams of aluminum sulfate is approximately 0.02 moles.
The molar mass of aluminum sulfate is 342.15 g/mol. Therefore, the mass of 0.25 moles of aluminum sulfate would be 85.54 grams (0.25 moles x 342.15 g/mol).
To determine the limiting reactant, we need to compare the moles of each reactant. First, calculate the moles of aluminum and copper sulfate separately. Then, determine the mole ratio between them and see which reactant is present in lower amount compared to the stoichiometric ratio. The reactant that is present in lower moles is the limiting reactant.
Well...Since Aluminum is insoluble in water as an element I assume you mean that an 0.12 mol Aluminum ions react with 0.12 mol Sulfate ions to form x mol of Aluminum Sulfate. Balanced equation (Kinda since since the other portions of the compounds are unknown): 2 Al3+(aq) + 3 SO42-(aq) <----->Al2(SO4)3 (aq) 0.12mol*(1mol Al2(SO4)3 (aq)/2mol Al3+)=0.06mol Al2(SO4)3 (aq) 0.12mol*(1molAl2(SO4)3/3mol SO42-) =0.04mol Al2(SO4)3 (aq) *Answer*
There are 5 moles of sulfur in 5 moles of H2SO4, as there is 1 mole of sulfur in each mole of H2SO4.
There are 3 moles of sulfate ions (SO4^2-) present in 1 mole of Al2(SO4)3. Therefore, in 1.7 moles of Al2(SO4)3, there would be 3 * 1.7 = 5.1 moles of sulfate ions.
4.2 moles of CS2 contain 8,4 moles sulfur.
To determine the number of moles of sulfur (S) present in 7.71g of S, you should first find the molar mass of sulfur, use it to convert the grams of sulfur to moles. Sulfur has a molar mass of approximately 32.06 g/mol, so the number of moles of S present in 7.71g is 7.71g / 32.06 g/mol = 0.24 moles of S.