4 Zn(s) + 10 HNO3(aq) 4 Zn(NO3)2(aq) + NH4NO3(aq) + 3 H2O(l)
Calculate the number of grams of zinc that must react with an excess of HNO3 to form 24.1 g NH4NO3.
the answer is expressed in G
2006-12-10 07:30:36 · 3 answers · asked by Jackson G 1 in Science & Mathematics ➔ Chemistry
get the atomic mass of Zn, write that above the Zn in the equation.
get the molecular mass of NH4NO3; write that above the NH4NO3 in the equation
Look at the coefficients in front of the Zn and the NH4NO3. Multiply the masses you just found by those coefficients.
(4 x Zn and 1 x NH4NO3)
This gives you the gram raio between Zn and NH4NO3. Use that raio to calculate grams of Zn needed to produce 24.1 g NH4NO3
24.1 g NH4NO3 x grams Zn/grams NH4NO3 = grams Zn
2006-12-10 07:40:03 · answer #1 · answered by The Old Professor 5 · 0⤊ 0⤋
24.1 g of NH4NO3, divided by 92.07 grams per mole is 0.262 moles of NH4NO3. the mole ratio between Zn and ammonium nitrate is 4, so times 0.262 moles by four to find the moles of Zn (1.048 moles). then multiply that by Zn's atomic mass, 65.41, and your answer is:
68.55 grams Zn
2006-12-10 07:38:18 · answer #2 · answered by car of boat 4 · 0⤊ 0⤋
maybe i'm reading it incorrectly, but i'm fairly sure this is wrong, it seems to be a combination of two reactions, there is an infinite number of ways to balance it. and i see you wrote the equation out, but you haven't balanced it. if you can do that, the actual stoichiometry is fairly straightfoward.
2016-05-23 02:39:32 · answer #3 · answered by Rebecca 4 · 0⤊ 0⤋