Calculate the volume of carbon dioxide produced from .025 g of aluminum carbonate and excess HCl at 25 degrees c and 745 toor according to the following reaction:
Al(2)(CO(3)(3)) (aq) + HCl (aq) ----- AlCl(3) (aq) + H(2)O + CO(2) (g)
im using the numbers in () to show a number that is really little beside the numbers.
PV=nRT
P=745 Torr/760 Torr = 0.98atm
V=?
n= 0.025g of Al/ 26.98g/mol of Al
R=0.0821
T=298K
V= (0.0009266*0.0821*298)/0.98
V=0.0231L