posted by Sammy on .
If 430 mL of 0.837 M aqueous HNO3 and 23.5 g of solid Al are reacted stoichiometrically according to the balanced equation, how many grams of solid Al remain? Round your answer to 3 significant figures.
6HNO3(aq) + 2Al(s) → 3H2(g) + 2Al(NO3)3(aq)
Limiting reagent problem are about th same from one to another.
1. Write and balance the equation.
6HNO3 + 2Al ==> 2Al(NO3)3 + 3H2
2. Convert HNO3 to moles. moles = M x L.
3. Convert g Al to moles. moles = g/molar mass.
4a. Using the coefficients in the balanced equation, convert moles HNO3 to moles of either product.
4b. Same procedure, convert moles Al to the same product you chos in 4a.
4c. Likely, these to moles will not be the same. Obviously, one of them is not correct. The correct is ALWAYS the smaller one. That identifies the limiting reagent.
5. IF you wanted to know grams H2 or grams Al(NO3)3 you would convert o gram by grams = moles x molar mass. That isn't in this problem.
6. You want the moles of the excess reagent. Obtain that this way.
a. Using the coefficients in the balanced equation (same way as 4a and 4b above) convert moles of the limiting reagent (now that it is identified) to the OTHER reactant (the one that you now know is the one in excess).
b. Convert mols of the excess reagent to grams. grams = moles x molar mass.
c. You know how much you started with. The last step tells you how much was used. The difference will tell you how much is left.