Sulfur dioxide gas (so2) and oxygen gas (o2) react to form the liquid product of sulfur trioxide (so3). how much sulfur dioxide would you need to completely react with 6.00 g o2 such that all reactants could be consumed?

1) Balanced chemical equation:

2SO2 (g) +  O2 (g) -> 2SO3 (l)

2) Molar ratios

2 mol SO2 : 1 mol O2 : 2 mol SO3

3) Convert 6.00 g O2 to moles

number of moles = mass in grams / molar mass

number of moles = 6.00 g / 32 g/mol = 0.1875 mol O2.

4) Use proportions with the molar ratios

=> 2 moles SO2 / 1 mol O2 = x / 0.1875 mol O2

=> x = 0.1875 mol O2 * 2 mol SO2 / 1 mol O2 = 0.375 mol SO2.

5) Convert 0.375 mol SO2 to grams

mass in grams = number of moles * molar mass

molar mass SO2 = 32 g/mol + 2*16 g/mol = 64 g/mol

=> mass SO2 = 0.375 mol * 64 g / mol = 24.0 g

24.0 g of SO2 are needed to react completely with 6.00 g O2.

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na