Posted by robb on Monday, March 30, 2009 at 6:36pm.
Hmm.. the answer should be the limiting reaction
We already know the grams of Zn being used, but we have to find the grams of H3PO4 being used.
(50.0gZn) * (1moleZn/65.39gZn) * (97.9953gH3PO4/1moleZn) = 74.93g H3PO4
Now you have to find the limiting reaction
(50.0gZn) * (1moleZn/65.39gZn) * (3moleH2/3moleZn) * (2.016gH2/1moleH2) = 1.54g H2
Do that same thing with the H3PO4 using the mass
(74.93gH3PO4)*(1moleH3PO4/97.9953gH3PO4)*(3moleH2/2moleH3PO4)* (2.016gH2/1moleH2) = 2.312g H2
1.54g H2 will be produced. Sorry if I'm wrong, or my work is messy.
To Rob and Chopsticks.
Correct answer of 1.54 g H2 produced. Long way to get there.
First, Zn is the limiting reagent because there is no amount of H3PO4 given. Limiting reagent problems are those where BOTH reactants are given; then one must find the limiting reagent. Where just one reagent is given, it is always assumed that there is enough of the other, usually an excess, to use all of the grams of the reagent listed.
So moles Zn =
50 g x (1 mole Zn/65.39) = 0.7646.
Moles H2 = mols Zn x (3 moles H2/3 moles Zn) = 0.7646 x (1/1) = 0.7646 moles H2.
Grams H2 = 0.7646 moles H2 x (2.016 grams/1 mole H2) = 1.5415 which rounds to 1.54 to three significant figures. Technically, the 50 grams has only one s.f. (unless it was 50.0 or 50. and you just omitted the period or the other zero) but I doubt the instructor meant for you to round to 1 gram H2 gas.