Well, it has a lot to do with it since instructions for working a limiting reagent problem is just an expanded set of rules (some don't apply) when doing a simpler stoichiometry problem. I have been handy tonight with passing out answers I've given, especially for limiting reagent problems, because I got tired of typing the same thing over and over again (and for the same problem, yet). However, if you got caught in the crossfire, I apologize; but, you can use that set if you ignore the items that don't apply. I worked this problem for someone but here it is again.
a. Convert 66.6 g NH3 to moles. moles = grams/molar mass. Using the coefficients in the balanced equation, convert moles NH3 to moles F2. Now convert moles F2 to grams using g = moles x molar mass.
b. Convert 5.65 g HF to moles. Using the coefficients in the balanced equation, convert moles HF to moles NH3. Now convert moles NH3 to grams. (See the similarity?)
c. Convert 225 g F2 to moles. Using the coefficients in the balanced equation, convert moles F2 to moles N2F4. Convert to grams.
what do limiting reagents have to do with...
5F2(g) + 2NH3(g) ==> N2F4(g) + 6HF(g)
a)If you have 66.6 g NH3, how many grams of F2 are required for complete reaction?
b)How many grams of NH3 are required to produce 4.65g HF?
c)How many grams of N2F4 can be produced from 225g F2?
1 answer
1 answer