Answer: 83%
Explanation:
The detailed solution is shown in the image attached. First we must work out the balanced reaction equation because accurate solution of the problem must be based on the stoichiometry of the reaction. From the given concentration and volume of reactants, we calculate the amount of substance reacted hence identify the limiting reactant. Lastly we use simple proportion to obtain the theoretical yield of the precipitate. This is now used to calculate the actual yield as shown in the solution attached.
Answer:
Explanation:
To find the concentration; let's first compute the average density and the average atomic weight.
For the average density 
; we have:
The average atomic weight is:
So; in terms of vanadium, the Concentration of iron is:
From a unit cell volume 
where;
= number of Avogadro constant.
SO; replacing with 
; with 
; 
with 
and
with 
Then:
![a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2FA_%7BFe%7D%20%5D%20%2B%20%5BC_v%2FA_v%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%5CBig%20%28%5Cdfrac%7B100%7D%7B%5B%28100-C_v%29%2F%5Crho_%7BFe%7D%20%5D%20%2B%20%5BC_v%2F%5Crho_v%5D%7D%20%5CBig%29%20%20%7D)
![a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100-C_v%29A_%7Bv%7D%20%5D%20%2B%20%5BC_v%2FA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100-C_v%29%2F%5Crho_%7Bv%7D%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
![a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }](https://tex.z-dn.net/?f=a%5E3%20%3D%20%5Cdfrac%20%20%20%7B%20n%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20A_%7BFe%7D%20%5Ctimes%20A_v%7D%7B%5B%28100A_%7Bv%7D-C_vA_%7Bv%7D%29%20%5D%20%2B%20%5BC_vA_Fe%5D%7D%20%5CBig%29%20%7D%20%20%20%20%7BN_A%20%20%5CBig%20%28%5Cdfrac%7B100%20%5Ctimes%20%5Crho_%7BFe%7D%20%5Ctimes%20%20%5Crho_v%20%7D%7B%5B%28100%5Crho_%7Bv%7D%20-%20C_v%20%5Crho_%7Bv%7D%29%20%5D%20%2B%20%5BC_v%20%5Crho_%7BFe%7D%5D%7D%20%5CBig%29%20%20%7D)
Replacing the values; we have:
(2) is the only correct answer. (1) 5 grams of H2 is 2.5 mols of this gas, which would result at STP 22.4 * 2.5 Liters (according to Avogadro's law). 5 moles of O2 would result in 5 * 22.4 Liters, and (4) 5 * 10^23 molecules of CO2 (which is what I assume you meant) would be 5 mols, which would result in 5* 22.4 Liters.
(2) is your answer
