4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous salt).
5,7 moles (SO4)3-.
1,125 moles of sodium sulfate contain 6,774908464125.10e23 molecules.
You need the molar solution to get the number of moles present in 6.52g of Zinc Sulfate.
25,3 moles of potassium sulfate hva a mass of 4,4409 kg.
There are 6 moles of sulfur present in 3 moles of aluminum sulfate, because aluminum sulfate has a 2:3 ratio of aluminum to sulfur.
4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous salt).
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 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.
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*
5,7 moles (SO4)3-.
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.
1,125 moles of sodium sulfate contain 6,774908464125.10e23 molecules.
To determine the number of moles of aluminum present, we need to first determine the molar mass of aluminum, which is approximately 26.98 g/mol. We can then use the formula: moles = mass / molar mass. Plugging in the values, we get moles = 15 g / 26.98 g/mol ≈ 0.56 moles of aluminum.
In aluminum sulfate, the molar mass of aluminum is 27 g/mol. Calculate the amount of aluminum in 5.60 g of aluminum sulfate using the molar ratio between aluminum and aluminum sulfate (1:1). Therefore, there are 5.60 grams of aluminum in 5.60 grams of aluminum sulfate.
To find the amount of aluminum needed to produce aluminum sulfate, you need to consider the molar mass of aluminum sulfate and the ratio of aluminum in the compound. First, calculate the molar mass of aluminum sulfate (Al2(SO4)3). Then, find the ratio of aluminum in the compound (2 moles of Al in 1 mole of Al2(SO4)3). Finally, use this information to calculate the grams of aluminum needed to produce 25.0 grams of aluminum sulfate.
To determine the number of moles of aluminum present in 856g, you need to divide the mass by the molar mass of aluminum. The molar mass of aluminum is approximately 26.98 g/mol. So, 856g ÷ 26.98 g/mol ≈ 31.7 moles of aluminum are present in 856g.