A2.00-l reaction vessel, initially at 298 k, contains chlorine gas at a partial pressure of 337 mmhg and fluorine gas at a partial pressure of 729 mmhg. identify the limiting reactant and determine the theoretical yield of clf3 in grams.

The theoretical yield of is 2.24 g

Explanation:

The equation given by ideal gas follows:

         .......(1)

where, P = pressure of the gas

V = Volume of the gas

T = Temperature of the gas

R = Gas constant =

n = number of moles of gas

We are given:

Putting values in equation 1, we get:

We are given:

Putting values in equation 1, we get:

The chemical equation for the reaction of chlorine gas and fluorine gas follows:

By Stoichiometry of the reaction:

3 moles of fluorine gas reacts with 1 mole of chlorine gas.

So, 0.0363 moles of fluorine gas will react with = of chlorine gas

As, given amount of chlorine gas is more than the required amount. So, it is considered as an excess reagent.

Thus, fluorine gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of fluorine gas produces 2 moles of

So, 0.0363 moles of fluorine gas will produce = of

Molar mass of = 92.45 g/mol

Moles of = 0.0242 moles

Putting values in above equation, we get:

Hence, the theoretical yield of is 2.24 g

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

the answer would be a