To prepare 47.64 g of nitric acid using 450 L of NO2 at 5.00 atm and 295 K, 0.762 moles of HNO3 can be produced.
To determine the number of moles of nitric acid that can be prepared using 450 L of NO2 at 5.00 atm and 295 K, we need to use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for n, which is the number of moles of NO2 gas.
First, we need to convert the given pressure and temperature to the appropriate units. 5.00 atm is equivalent to 504.336 kPa, and 295 K is equivalent to 268.15 K.
Using the ideal gas law equation, we can solve for n:
n = PV/RT = (504.336 kPa)(450 L)/(0.0821 L·atm/mol·K)(268.15 K) = 0.254 mol NO2
According to the balanced chemical equation, 3 moles of NO2 react with 1 mole of H2O to produce 2 moles of HNO3 and 1 mole of NO gas. Therefore, we need 3 times the number of moles of NO2 to produce nitric acid:
n(HNO3) = 3 × n(NO2) = 3 × 0.254 mol = 0.762 mol
Finally, we can use the molar mass of HNO3 to convert moles to grams:
m(HNO3) = n(HNO3) × M(HNO3) = 0.762 mol × 63.01 g/mol = 47.64 g
Therefore, 47.64 g of nitric acid can be prepared using 450 L of NO2 at 5.00 atm and 295 K.