2. Write the equilibrium constant expression for Al2(SO4)3 and look up the value for the Ksp. 3. Calculate the molar solubility and the solubility of Al2(SO4)3. 4. A chemist adds Al2(SO4)3 solid to water. The calculations you performed in number 3 above apply to this initial solution. 100 mL of a 0.10-M AlCl3 solution is mixed in.

Al has a 3 + charge, and SO 4 has a 2-charge. So, it takes 3 of the SO 4 anions to balance 2 of the Al cations. So, you get Al 2 (SO 4) 3. Now, for naming this compound, since Al is ONLY 3 + and does not have variable valence, you don't use the (III) in parentheses. The Roman number in parentheses is ONLY used for transition metals that have ...

Question: Al2(SO4)3(aq) + AgNO3(aq) → Express your answer as a chemical equation. Enter NOREACTION if no reaction occurs.

2021年12月20日 · 2 Al + 3FeSO4 → Al2(SO4)3 + Fe. look at oxidation number (o.n.) for each element on left side and right side of the equation . Al on left is zero. Al on right is 3+. Al has lost electrons so it was oxidized. It is the reducing agent. Fe on the left is 2+. Fe on the right is zero. Fe has gained electrons so it was reduced.

2020年11月20日 · 2Al(s) + 3H2SO4(aq) ==> Al2(SO4)3(aq) + 3H2(g) ... balanced equation. Using the stoichiometry of the balanced equation (2 mol Al : 3 mol H 2 SO 4) and dimensional analysis, we can find the mass of Al needed to react with 26.5 ml of 0.542 M H2SO4. How many moles H2SO4 are present? 26.5 ml x 1 L/1000 ml x 0.542 mol/L = 0.014363 moles H 2 SO 4

2023年5月24日 · Haroon I. To determine the total concentration of all ions in a solution formed by dissolving 1.50 g of Al2(SO4)3 in 100.0 ml of solution, we need to calculate the moles of Al2(SO4)3 and then determine the concentration of each ion.

2017年8月14日 · The only reason you are "just worried about the compounds with Al" is because it turns out that Al2(SO4)3 is the limiting reactant in this problem. Therefore the moles of this reactant will dictate the moles of all products formed.

2021年2月23日 · We see that in this case, when we have 6 moles of NaOH in the reactant, it goes on to form 2 moles of Al(OH) 3 in the products. Thus the molar ratio can be written as follows: mole ratio = the moles of NaOH/the moles of Al(OH) 3 = 6 moles NaOH/ 2 moles Al(OH) 3 = 3 moles NaOH/1 mole Al(OH) 3

2021年1月8日 · We need the molecular weights of both the Al 2 (SO 4) 3 and sulfur, S. Al 2 (SO 4) 3 = 342.2 amu. S = 32 amu. We have 3 sulfurs per molecule, so the total contribution of sulfur to the 342.2 is 3 x 32 = 96 amu. The percentage is thus (96amu)/(342.2 amu) = 0.28 or 28%. Bob

How many grams of Al2(SO4)3 are generated when 238 g of H2SO4 reacts? 277 g of Al2(SO4)3 2,490 g of Consider the reaction: 2 Al(OH) 3 + 3 H 2 SO 4 → Al 2 (SO 4 ) 3 + 6 H 2 O.