Homework Help: chemistry

Posted by lyne on Tuesday, January 27, 2009 at 8:42pm.

a.Write the balanced chemical equation for the reaction of methane and sulfur. (Type your answer using the format CO2 for CO2. Use the lowest possible coefficients.)

1CH4 + 4S --> 1CS2 + 2H2S

(b) Calculate the theoretical yield of carbon disulfide when 240. g of methane is reacted with an equal mass of sulfur.

i got a. how do i do b.?

chemistry - DrBob222, Tuesday, January 27, 2009 at 8:52pm
Do you know how to do limiting reagent problems?
Convert 24.0 g CH4 to moles.
Convert 24.0 g S to moles.
Convert moles CH4 to moles CS2 and convert moles S to moles CS2 and take the smaller number of moles CS2. The one producing that (CH4 or S) will be the limiting reagent. Then convert moles CS2 to grams. That will be the theoretical yield.
chemistry - lyne, Tuesday, January 27, 2009 at 8:55pm
oh okay yea that sounds familiar!
so is it the same process for this...

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is given below.
Fe2O3(s) + 2 Al(s) 2 Fe(l) + Al2O3(s)

What masses of iron(III) oxide and aluminum must be used to produce 28.0 g iron?
chemistry - DrBob222, Tuesday, January 27, 2009 at 9:08pm
Yes, and no. This is not a limiting reagent problem.
Just convert 28.0 g iron to moles, convert mols Fe to moles Fe2O3(using the coefficients in the balanced equation), and convert that to grams Fe2O3.
Then do a separate problem, converting 28.0 g Fe to moles, convert that to moles Al, then convert that to grams Al
chemistry - lyne, Tuesday, January 27, 2009 at 9:14pm
okay so i got 40.0g Fe2O3 and 13.5g Al
how do i use this to get the last part
chemistry - DrBob222, Tuesday, January 27, 2009 at 9:47pm
Is there a last part you didn't post? The answer of 13.5 g and 40.0 g Fe2O3 IS the amount of Al and Fe2O3 required to produce 29.0 g Fe which is what the question asked. There was no other part to it that I see.
chemistry - lyne, Tuesday, January 27, 2009 at 9:51pm
yea sorry my bad i didnt post it but i figured it out :)

thanks for all the help

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