The combustion reaction of acetylene has a balanced equation
C2H2+3O2--->2H2O+2CO2
Using the ratio 1:3:2:2, if 1.10 mols of CO2 are produced, assuming complete combustion, than that means that 0.55 mols of acetylene (C2H2) must have been combusted. The molar mass of acetylene is 26, so the answer is 26*0.55 = 14.3 grams
Answer:
The answer to your question is: 0.118 moles of water
Explanation:
Data
moles of water = ?
Vol of O2= 1.6
Temp = 324°K
Pressure = 0.984 atm
R = 0.082 atm l / mol°K
Process
1.- Calculate the moles of O₂ using the ideal gas formula.
PV = nRT
n = 0.059 moles of oxygen
2.- Find the moles of water
2 moles of water ------------------ 1 mol of oxygen
x ------------------ 0.059 moles
x = (0.059 x 2) / 1
x = 0.118 moles of water
Answer:
N=35%. so 35/100*7 =2.45%
H=5.0 so 5/100*1=0.05%
o=60% so 60/100*8=4.8%
Answer:
I am pretty sure Danny Duncan told me 69
Explanation:
niice
you would physically separate sand and water