|By Joel_Set (Joel_Set) on Monday, October 06, 2003 - 06:24 am: Edit|
Okay I have a few questions. If anyone can answer and provide a thorough explanation, I would really appericiate it. Shall we cut the crap and move on?
1. Nitroglycerine decomposes violenetly according to the unbalanced chemical equation below. How many total moles of gases are produced from the decomposition of 1.00 mole C3H5(NO3)3?
C3H5(NO3)3--> CO2 +N2 + H20 + 02 (not balanced)
2. Amixture of MgCO3 and MgCO3 * 3H20 has a mass of 2.883g. After heating to drive off all the water the mass is 1.927g. What is the mass percent of MgCO3*3H20 in the mixture?
|By Y17k (Y17k) on Monday, October 06, 2003 - 08:10 am: Edit|
2. Molar mass of MgCO3 = 84.32 g/mole
2.883 - 1.927 = .9560 grams of water
Molar mass of H20= 18.02 g/mole
so, n(H20) = .9560/18.02 = 0.05305 mols
.05305/3 = .01768 mols of MgCO3 were hydrated.
Mass of MgCO3 that were hydrated = 84.32 * .01768 = 1.491 grams
1.491 + .956 = 2.447 grams of MgCO3*3H2O
% yield = 2.447/2.883 *100 = 84.88 %
phew... i gotta review stiochiometry again ><
|By Fairyofwind (Fairyofwind) on Monday, October 06, 2003 - 08:31 am: Edit|
lol the font almost seems like a ligature "hornies..."
|By Joel_Set (Joel_Set) on Monday, October 06, 2003 - 09:03 am: Edit|
lol myabe i did put homies..
Thx y17k for the 2nd prob..
can anyone do the 3rd?
|By Y17k (Y17k) on Monday, October 06, 2003 - 09:56 am: Edit|
oops i skipped the first problem
btw did i get the 2nd one right?
1. 4 C3H5(NO3)3 ----> 12 CO2 + 6 N2 + 10 H20 + 02
the gases are CO2, N2, and O2
therefore total number of mols of gases FOR 4 moles of C3H5(NO3)3 = 12+6+1 = 19
for ONE mole, 19/4 = 4.75 mols of gas
|By Joel_Set (Joel_Set) on Tuesday, October 07, 2003 - 08:21 am: Edit|
yeah u got the second one right.. i had it once but i just had forgotten how i got it.. :/
i think the second one is right too.. wow ur good at chemistry thx..
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