Copper(II) ions can be displaced by aluminum metal.
3 CuSO4(aq) + 2 Al(s) → Al2(SO4)3(aq) + 3 Cu(s)
How many grams of Al would be needed to displace all of the copper ions in 40.5 mL of 0.420 M CuSO4?
How many grams of Cu would be recovered from the copper ions in 40.5 mL of 0.420 M CuSO4?
Copper(II) ions can be displaced by aluminum metal. 3 CuSO4(aq) + 2 Al(s) → Al2(SO4)3(aq) +...
N THE FOLLOWING INFORMATION: 3 Cuso, (aq) → 3 Cu (s) + A12(SO4)3 (aq) GIVE 6.75 grams of Aluminum Density of aluminum- 2.70 g/mL Tolarity of the CuSO, solution- 0.345 M CALCULATE: A. Volume of the Al (s) sample Mass in grams of copper produced C. Volume in mL of the CuSO4 (aq) solution used D. Mass in grams of the product Al(SO)s produced
Sulfuric acid (H2SO4) dissolves Aluminum metal according to the reaction: 2 Al(s)+ 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g) Suppose you want to dissolve an Aluminum block with a mass of 15.2 g. What minimum mass of H2SO4 (in g) do you need? What mass of H2 gas (in g) can the complete reaction of the aluminum block produce? Please Show Work
Aluminum metal reacts with sulfuric acid according to the following equation: 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(s) + 3H2(g) If 12.9 g of aluminum reacts with excess sulfuric acid, and 62.4 g of Al2(SO4)3 are collected, what is the percent yield of Al2(SO4)3?
a) How many mol of Al2(SO4)3 are required to make 40 mL of a 0.050 M Al2(SO4)3 solution? (Hint, molarity is moles per liter. You know the molarity. You are given the number of mL's, which can be converted to liters. Just set it up so the units cancel and you get "moles".) moles Al2(SO4)3 b) What is the molecular weight of Al2(SO4)3, to the nearest gram? c) What mass of Al2(SO4)3 is required to make 40 mL of a 0.050 M...
a) How many moles of Al2(SO4)3 are required to make 44 mL of a 0.090 M Al2(SO4)3 solution? (Hint, molarity is moles per liter. You know the molarity. You are given the number of mL's, which can be converted to liters. Just set it up so the units cancel and you get "moles".) moles Al2(SO4)3 b) What is the molecular weight of Al2(SO4)3, to the nearest gram? grams c) What mass of Al2(SO4)3 is required to make 44 mL of...
Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al(s)+3H2SO4(aq)-----=Al2(SO4)3(aq)+3H2(g) Suppose you wanted to dissolve an aluminum block with a mass of 14.9g . What minimum mass of H2SO4 would you need? What mass of H2 gas would be produced by the complete reaction of the aluminum block? Express your answer in grams.
a) How many moles of Al2(SO4)3 are required to make 55 mL of a 0.050 M Al2(SO4)3 solution? (Hint, molarity is moles per liter. You know the molarity. You are given the number of mL's, which can be converted to liters. Just set it up so the units cancel and you get "moles".) moles Al2(SO4)3 b) What is the molecular weight of Al2(SO4)3, to the nearest gram? grams c) What mass of Al2(SO4)3 is required to make 55 mL of...
8. Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g) Suppose you wanted to dissolve an aluminum block with a mass of 15.9 g. [0.5] a. What minimum mass of H2SO4 would you need? b. What mass of H2 gas would be produced by the complete reaction of the aluminum block?
Aluminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3(s)+3H2SO4(aq)→Al2(SO4)3(aq)+6H2O(l) a. Which reagent is the limiting reactant when 0.450 mol Al(OH)3 and 0.450 mol H2SO4 are allowed to react? b. How many moles of Al2(SO4)3 can form under these conditions? c. How many moles of the excess reactant remain after the completion of the reaction?
E: How many moles of aluminum ions are present in 0.32 moles of Al2(SO4)3? F: How many moles of sulfate ions (SO42?) are present in 1.3 moles of Al2(SO4)3? G: Calculate molar mass C6H14O6 (sorbitol) Express your answer to two decimal places and include the appropriate units. H: The sedative Demerol hydrochloride has the formula C15H22ClNO2. How many grams are in 0.025 moles of Demerol hydrochloride? I: How many moles of H2 are needed to react with 0.55 mol of...