For the reaction
2Al+3H2SO4⟶3H2+Al2(SO4)32Al+3H2SO4⟶3H2+Al2(SO4)3
how many grams of sulfuric acid, H2SO4, are needed to react completely with 72.1 g of aluminum, Al?
For the reaction 2Al+3H2SO4⟶3H2+Al2(SO4)32Al+3H2SO4⟶3H2+Al2(SO4)3 how many grams of sulfuric acid, H2SO4, are needed to react completely...
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.
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?
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?
2Al(OH)3 + 3H2SO4 -----> Al2(SO4)3 + 6H2O In the above reaction, how many grams of Al(OH)3 would be consumed by the reaction of 3.8 moles of H2SO4 ?
2Al+3H2SO4⟶3H2+Al2(SO4)3 how many moles of hydrogen will be produced from 17.6 g of aluminum?
Aluminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3(s)+3H2SO4(aq)→Al2(SO4)3(aq)+6H2O(l) Which reagent is the limiting reactant when 0.550 mol Al(OH)3 and 0.550 mol H2SO4 are allowed to react? Al2(SO4)3(aq) Al(OH)3(s) H2O(l) H2SO4(aq)
2Al(OH)3 + 3H2SO4 -----> Al2(SO4)3 + 6H2O In the above reaction, how many moles of Al2(SO4)3 would be produced by the reaction of 3.8 moles of H2SO4 ?
How many grans of aluminum hydroxide will react with 75.0mL of 0.319M of sulfuric acid solution? 3. Aluminum hydroxide will react with sulfuric acid to yield aluminum sulfate and water according to the equation below: (4 pt) 3H2SO4(aq) + 2Al(OH)3(aq) + Al(SO4)3(aq) + 6H2O(1) a) How many grams of aluminum hydroxide will react with 75.0 mL of 0.319 M sulfuric acid solution?
How many g Al will react with 24.0 mL of 1.22 M H2SO4? 2Al (s) + 3H2SO4 (aq) → Al2(SO4)3 (aq) + 3 H2 (g)