Thread: Calculate the volume of acid.

2Al + 6HNO3 -------> 2Al(NO3)3 + 3H2

What volume of 2.40M acid ( HNO3) is required to have a complete reaction with the mass of Al ( 39.1g & that is about 1.45 mol ).

Here is what I did but I am not sure if it is right

# mol acid/ 1.45 mol Al = 6 mol acid/ 2 mol Al

#mol acid = 1.8125

now if that's right, what do i do next to find the volume?

Thank you in advance!

Yes, you need 6/2= 3 times as many mols of acid as of aluminum. Since you have 1.45 mols of aluminum, you will need 3(1.44)= 4.35 mols of acid. Now by "2.40M acid", I presume you mean one mol of acid per liter (it's been a while since I took chemistry!) so you would need liters of acid. That is volume, not number of mols.

oh great thank you!