Al(OH)3+3CH3COOH= Al(CH3COO)3 + 3H20 using the equation, determine the mass of aluminum acetate

aluminum-hydroxide
acetic-acid

#1

Al(OH)3+3CH3COOH= Al(CH3COO)3 + 3H20 using the equation, determine the mass of aluminum acetate that can be made if i do this reaction with 125 grams of acetic acid and 275 grams of aluminum hydroxide?

Answer:

I’ll just tell you the necessary steps to solve this problem since I don’t feel like doing calculations now.
-First you have to convert 125g and 275g to moles of each substance.
-Once you do that you need to find the limiting reactant between the two. When you know which one is the limiting reactant use simple stoichiometry to find moles of product (Al(CH3COO)3).
-Either calculate the molar mass or look it up online and multiply the molar mass by the amount of moles, then it will give you the grams of product asked.