Question

For the reaction, Ra + 2 HF --> RaF2 + H2, calculate the percent yield of radium fluoride if 300 grams of radium react with excess hydrofluoric acid (HF)

to yield 295 g of radium fluoride.

Answer 1

Y=84.2%

Step-by-step explanation:

Hello there!

In this case, for this stoichiometry problem, we need to consider the given balanced chemical reaction, in order to calculate the theoretical yield of radium fluoride, according to the stoichiometry and the 1:1 mole ratio between them:

`m_(RaF_2)^(theoretical)=300gRa*(1molRa)/(226.03g) *(1molRaF_2)/(1molRa)*(264.02gRaF_2)/(1molRaF_2) \\\\m_(RaF_2)^(theoretical)=350.4gRaF_2`

Finally, given the actual yield of the product, we can obtain the percent yield by dividing them:

`Y=(295g)/(350.4g) *100\%\\\\Y=84.2\%`

Best regards!