the big number describes the number ratio in a chemical equation
so for example,
2H2 + O2 --> 2H2O means
2 moles of hydrogen reacts with one mole of oxygen to form 2 moles of water
and as you know, the small (subscript) number determines the number of atoms of that element in one molecule of a compound
so I believe that drawing a normal lewis structure ( O=O ) should be correct
Gasoline would be a mixture if I recalled
<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
Use higher concentration of acid to produce 2 moles of hydrogen to form water
The reaction is
CaC₂(s) + 2H₂O (l) -----> Ca(OH)₂ (s) + C₂H₂ (g)
As we have data of gas ethyne (or acetylene), C₂H₂
We can calculate the moles of acetylene and from this we can estimate the mass of calcium carbide taken
the moles of acetylene will be calculated using ideal gas equation
PV =nRT
R = gas constant = 0.0821 Latm/molK
T = 385 K
V = volume = 550 L
P = Pressure = 1.25 atm
n = moles = ?
n = PV /RT = 1.25 X 550 / 0.0821 X 385 = 21.75 mol
As per balanced equation these moles of acetylene will be obtained from same moles of calcium carbide
moles of calcium carbide = 21.75mol
molar mass of CaC₂ = 40 + 24 = 64
mass of CaC₂ = moles X molar mass = 21.75 X 64 = 1392g
