Asuume the folowing molecular weight: MArshmellow :7.50 g/mol, Graham Cracker :

Question

Asuume the folowing molecular weight: MArshmellow :7.50 g/mol, Graham Cracker : 7.7g/mol , Chocolate 14g/mol , Smore 37.oo g/mol

Given 45 g of Marshmellow, 50 g of Graham Cracker and 70 g of Chocolate so how many moles
can you make? what would be theoretical yield in grams?
how much in grams of the excess reagents are left over?

Explanation / Answer

the usual steps for this type of problem. ie memorize them... 1) write a balanced equation 2) convert everything to moles 3) determine limiting reagent 4) convert moles limiting reagent to moles of desired species 5) convert moles to mass.. this is theoretical yield. *** 1 *** Fe(NO3)3(aq) + Na3(PO4)(aq) -----> FePO4(s) + 3 NaNO3(aq) iron(III) phosphate is insoluble. sodium nitrate is soluble. means FePO4 is the product. --> note.. I assumed sodium phosphate was Na3(PO4).. if you meant Na2H(PO4) or NaH2(PO4), you will need to adjust. *** 2 *** mw Fe(NO3)3 = 241.9 g/mole mw Na3(PO4) = 163.9 g/mole. moles Fe(NO3)3 = 21.14 g x (1 mole / 241.9 g) = 0.08739 moles moles Na3(PO4) = 11.55 g x (1 mole / 163.9 g) = 0.07047 moles *** 3 *** from balanced equation, 1 mole Fe(NO3)3 reacts with 1 mole Na3(PO4)... so... 0.08739 moles Fe(NO3)3 reacts with 0.08739 moles Na3(PO4), since we don't have that many moles of Na3(PO4), Na3(PO4) is the limiting reagent. *** 4 *** from balanced equation 1 mole Na3(PO4) ---> 1 mole FePO4(s) 0.07047 moles Na3(PO4) ----> 0.07047 moles FePO4(s) *** 5 *** finally... mw FePO4(s) = 150.8 g/mole 0.07047 moles FePO4 x (150.8 g / mole ) = 10.63 g FePO4.. 4 sig figs

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