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!