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Chemistry 27 Online
OpenStudy (anonymous):

Given the following chemical equation, if 50.1 grams of silicon dioxide is heated with excess carbon and 32.3 grams of silicon carbide is produced, what is the percent yield for this reaction? SiO2 (s) + 3C (s) arrow SiC (s) + 2CO (g) Answer 97% 75% 103% 48% I believe this is 97, am I correct?

OpenStudy (aaronq):

can you show your work?

OpenStudy (anonymous):

hold on wrong work! lol

OpenStudy (aaronq):

haha ok. i just wanna see your thought process ..and that you're not actually just guessing.

OpenStudy (anonymous):

I thought it was 96 but checking back over my work I'm not sure - I think i made some errors. I know that the total mass of reactants = 60 + 36 = 96 grams

OpenStudy (anonymous):

Because of that I presumed it would be 97% as that's closed but I realize I need to calculate the percent of yield. @aaronq

OpenStudy (aaronq):

that's not how you do it.. at all :P you need to convert the amounts in grams given to moles. moles=\(\dfrac{m}{M}\) then use the stoichiometric coefficients to build a ratio: SiO2 (s) + 3C (s) arrow SiC (s) + 2CO (g), \(\dfrac{n_{SiO_2}}{1}=\dfrac{n_{SiC}}{1}\) since they're both 1, we can just say that the moles are equal. SO, to find the percent yield, you divide the moles (or grams) actually produced, by the theoretical yield times 100%. percent efficiency = \(\dfrac{mass \;obtained}{theoretical\; mass}*100\%\)

OpenStudy (chmvijay):

did u get it ? @Pumpkin890

OpenStudy (chmvijay):

can u tell me molar mass of SIO2 and SiC

OpenStudy (anonymous):

Mass obtained? Would 50.1 be mass obtained and theoretical mass be 32.3?

OpenStudy (chmvijay):

nope :(

OpenStudy (anonymous):

60.08 g/mol = SIO2 40.11 g/mol SiC

OpenStudy (anonymous):

For the molar masses @chmvijay

OpenStudy (chmvijay):

ok now 60.08 gram SIO2 should produce 40.11 gram of SiC right according to equation but 50.1 gram of SiO2 should give = 40.11*50.1 /60.08= 33.44 gram of SIC but its forming only about 32.3 gram? can u solve now

OpenStudy (anonymous):

percent efficiency = 33.44/ 32.2 ∗100% Is that right?

OpenStudy (anonymous):

Ah! 103! Answer choice C Thank you @chmvijay

OpenStudy (chmvijay):

noo for 33.44 ----->100 for 32.3 gram=100*32.3/33.44 =

OpenStudy (anonymous):

oOH OKAY. one more time here..

OpenStudy (chmvijay):

LOL

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