Question

A student reacts 0.506 grams of aluminum with excess KOH and H2O according to reaction one from the lab manual. Enter your answers in numerical format How many moles of aluminum was reacted? _______ moles Al How many moles of K[Al(OH)4] will theoretically be produced? _______ moles K[Al(OH)4]

Reaction one is 2Al + 2KOH + 6H2O = 2K[Al(OH)4] + 3H2.

I have already tried converted 0.506 grams of Al to moles and got the answer 0.019, but it is still marked as incorrect.

Answer 1

Given reaction is

2 Al + 2 KOH + 6 H2O ---> 2 K[Al(OH)4] + 3 H2.

2 mole 2 mole 6 mole 2 mole 3 mole

Given

Mass of Al = 0.506 g

Molar mass of Al = 27 g/mol

No. of moles of Al = Mass / Molar mass = 0.506 g / 27 g/mol = 0.0187 moles

No. of moles of K[Al(OH)4] theortically produced = No. of moles of Al reacted = 0.0187 moles

Check this values sometimes no. of significant figures might be a problem

if it doesn't work post the question directly as photo there might be some other thing that you might have missed