The combustion of acetylene gas is represented by this equation:
2C2H2 + 5O2 --------> 4Co2 + 2H2O2
(sorry that subscripts don't work)
a) How many grams of CO2 and grams of H2O are produced when 52.0g C2H2 burns?
b)How many grams of oxygen are required to burn 52.0g C2H2?
c)Use the answeres from A and B to show that this equation obeys the law of conservation of energy.
MY ANSWERS
a)175.78g CO2
35.98g O2
211.76g total
b)6.14g O2
c) ?
WHAT I DID
* means multiply, / means fraction
a) How many grams of CO2 and grams of H2O are produced when 52.0g C2H2 burns?
5.20g C2H2 * (1 mol C2H2 / 26.04g C2H2) * (4mol CO2 / 2mol C2H2) * (44.01g Co2 / 1 mol CO2) = 175.78g CO2
52.0gC2H2 *( 1 mol C2H2 / 26.04g C2H2) * (2 mol H2O / 2 mol C2H2) * (18.02g H20 / 1 mol H20) = 35.98g O2
= 211.76g total
b) How many grams of oxygen are required to burn 52.0g C2H2?
52.0g C2H2 * (1 mol C2H2 / 26.04g C2H2) * (5 mol O2 / 2 mol C2H2) * (32.00g O2 / 1 mol O2) = 6.14g O2
2007-11-14 09:55:01 · 2 answers · asked by Emily A 3 in Science & Mathematics ➔ Chemistry
C is where I get confused. The law of conservation of mass states that what goes into the equation must come out the other side. When I add it all together, I get them 11 grams apart. Therefore, I did something wrong along the way. I already did the problem twice, and I can't figure it out.
2007-11-14 09:56:27 · update #1
THANK YOU!!! I was just adding it up wrong over and over again. THANK YOU SOO MUCH!!
2007-11-14 10:13:16 · update #2
and keep your answers coming. I need to understand this to my fullest. I have to present these problems in class tomorow.
2007-11-14 10:18:22 · update #3
check your math on B your set up is correct it should be around 160g
C) add up the masses of the reactants this should equal the mass of the products. That is the law of conservation of mass. No mass is lost.
2007-11-14 10:01:08 · answer #1 · answered by kentchemistry.com 7 · 0⤊ 0⤋
2C2H2 + 3O2 ===> 4CO2 +2H2O
Atomic weights: C=12 H=1 C2H2=26 O=16 CO2=44 H2O=18 O2=32
a) 52.0gC2H2 x 1molC2H2/26gC2H2 x 4molCO2/2molC2H2 x 44gCO2/1molCO2 = 176g CO2
52.0gC2H2 x 1molC2H2/26gC2H2 x 2molH2O/2molC2H2 x 18gH2O/1molH2O = 36g H2O
b) 52.0gC2H2 x 1molC2H2/26gC2H2 x 3molO2/2molC2H6 x 32gO2/1molO2 = 96g O2
2007-11-14 18:17:29 · answer #2 · answered by steve_geo1 7 · 0⤊ 0⤋