4 AlF3 + 3 O2 > 2Al2O3 + 6 F2 How many moles of AlF3 is needed to produce 8.25 moles of F2 from the given balanced equation shown above? (SHOW WORK)

Explanation:

6 F2>4 AlF3

F2> 4/6 AlF3

8.25 F2 > 4×8.25/6 AlF3

so 5.5 moles

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.