Question

calculate the mass of so3 producer is 500 grams of acid reacts with 242 according to the equation to SO2 + O2 gives to so3 identify the limiting reagent ​

Answer 1

Answer:

SO2 is the limiting reactant.

624.788g SO3

Explanation:

So first, we need to figure out the stoichiometric and feed ratios to determine the limiting reagent.

Reaction:

2SO2 + O2----> 2SO3

So, for the stoichiometric ratio, we are just going to use the stoichiometric coefficients of the reactants:

  $\frac{2 mol SO2}{1 mol O2} =\frac{2}{1}$

So for the feed ratio, we're gonna have to find the moles for the reactants in the question to find the feed ratio.

For So2:

$(500gSO2)(\frac{1molSO2}{64.066gSO2} ) =7.804mol SO2$

For O2:

$(242g O2)(\frac{1 mol O2}{15.999g O2} ) =15.126mol O2$

Now, feed ratio:

$\frac{7.804 mol SO2}{15.126 mol O2} =\frac{0.5159}{1}$

Feed ratio<Stoichiometric ratio.

Easier way to look at it is if mol O2  is kept constant the two ratios, the moles of SO2 are less in the feed are less that the ones requıred stoichiometrically. Hence SO2 is the limiting reactant.

To calculate mass of SO3 we use moles of SO2 i.e limiting reactant.

From equation:

$(\frac{2 mol SO3}{2 mol SO2} ) (7.804molSO2)=7.804 mol SO3$

To find mass of SO3:

$(7.804mol SO3)(\frac{80.06g SO3}{1 mol SO3})=624.788g SO3$