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 ?
2Al(OH)3 + 3H2SO4 -----> Al2(SO4)3 + 6H2O In the above reaction, how many moles of Al2(SO4)3...
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 ?
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?
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?
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)
Consider the reaction: 2 Al(OH)3 + 3 H2SO4 à Al2(SO4)3 + 6H2O. How many grams of Al2(SO4)3 are generated when 152 g of H2SO4 reacts? Show Work on Scratch Paper! O 530.g of Al2(SO4)3 O 1590 g of Al2(SO4)3 177 g of Al2(SO4)3 43.6 g of Al2(SO4)3 131 g 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?
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?
Mixing 39.0g of Al(OH)3 with an excess of H2SO4 we produce Al2(SO4)3. Using the following equation answer the following: I have already balanced the equation from Al(OH)3+H2SO4---->Al2(SO4)3+H2O 2Al(OH)3 + 3H2504 → Al2(SO4)3 + 6H2O a. Determine and circle the limiting reagent for A12(SO4)3 (show all its calculations). b. Determine and circle the yielding theory for Al2(S04)3 (show all its calculations). of Al2(SO4)3 (show all its Determine the % of yield, if the experimental yield is 73.25 calculations). c.