Consider the reaction: 2 so2(g)+o2(g)→2 so3(g) if 285.5 ml of so2 reacts with 158.9 ml of o2 (both measured at 315 k and 50.0 mmhg), what is the limiting reactant and the theoretical yield of so3? if 187.2 ml of so3 is collected (measured at 315 k and 50.0 mmhg), what is the percent yield for the reaction?

a. Oxygen is the limiting reagent.

b. %

Explanation:

Hello,

a. Limiting reagent and sulfur trioxide's theoretical yield.

At first, we must compute the involved moles for both sulfur dioxide's and oxygen's as follows, considering the volumes in liters and the pressure in atm of 50.0mmHg*1atm/760mmHg=0.0658atm:

Afterwards, by considering the properly balanced chemical reaction:

We compute the oxygen's moles that completely reacts with the previously computed moles of as follows:

That result let us know that the oxygen is the limiting reagent since just moles are available in comparison with the moles that completely would react with moles of .

Now, to compute the theoretical yield of sulfur trioxide, we apply the following stoichiometric relationship:

b. Percent yield.

At first, we must compute the collected (real) moles of sulfur trioxide:

Finally, we compute the percent yield:

%

%

%

Best regards.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g