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.
To solve this problem, we need to determine the amount of HNO3 that reacts with Al, and then subtract it from the initial amount of Al to find how much Al remains.
First, let's calculate the number of moles of HNO3 in the given solution. We have 430 mL of a 0.837 M (moles per liter) solution of HNO3. Since 1 L is equal to 1000 mL, we can convert the volume to liters by dividing by 1000:
430 mL ÷ 1000 mL/L = 0.43 L
Next, we can calculate the number of moles of HNO3 by multiplying the volume in liters by the molarity:
0.43 L × 0.837 mol/L = 0.36051 mol
Now, let's determine the stoichiometry of the balanced equation. From the balanced equation:
6 moles of HNO3 react with 2 moles of Al
So, the ratio of moles of HNO3 to moles of Al is 6:2, which simplifies to 3:1.
Therefore, if 0.36051 mol of HNO3 reacts, the number of moles of Al that reacts is 1/3 of that:
0.36051 mol × (1/3) = 0.12017 mol
The molar mass of aluminum (Al) is 26.98 g/mol. To find the mass of Al that reacts, we can multiply the number of moles by the molar mass:
0.12017 mol × 26.98 g/mol = 3.2418 g
Finally, to determine the mass of Al that remains, we subtract the mass that reacts from the initial mass of Al:
23.5 g - 3.2418 g = 20.2582 g
Rounding to three significant figures, the mass of solid Al remaining is approximately 20.3 g.