having trouble with this....
2Al(OH)3 + 3H2SO4 goes to Al2SO3 + 6H20
If 15.0g of Al(OH)2 and 20.0g of H2SO4 are mixed together, the number of grams of Al(OH)3 left unreated at the end of the reaction is;
So far i have only worked out the moles of Al(OH)3 is 0.1923
and the moles of H2SO4 is .20389
from there im stuck... thanks in advance for any help