2C6H14 + 13O2 ---> 6CO2 +14H2O
Calculating the molar mass of C6H14: M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol
Thus, 86.17 g of C6H14 corresponds to 1 mole.
2C6H14 + 13O2 ---> 6CO2 +14H2O
based on the equation 2 mol 6 mol
according to the question 1 mol 3 mol
To determine M(CO2): M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
Therefore, 3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Final answer: 132.0 g CO2
Answer:
Explanation:
We have:
V₁ = 18.5 L
T₁ = 18.5° C = 273 + 18.5 = 291.5 K
V₂ = 19.8 L
T₂ =?
Pressure remains constant
Applying the ideal gas law
CxHy + (x+0.25)O₂ → xCO₂ + 0.5yH₂O
m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}
0.2845/{44.01x}=0.1451/{9.01y}
x/y=0.4=2:5
The empirical formula is C₂H₅.
Response:
a. To purify a gypsum sample, you will need the following equipment: Bunsen burner, beaker, filter funnel, stirring rod, and filter paper.
b. Gypsum, a sulfate mineral consisting of calcium sulfate dihydrate, can be purified by following these steps:
1. Add water to the gypsum in a beaker.
2. Stir the mixture thoroughly with the stirring rod.
3. Use the filter paper and filter funnel to remove excess solids from the mixture.
4. Heat the filtered mixture on the Bunsen burner to evaporate the remaining water.
5. After cooling, filter again through the filter paper to obtain pure gypsum.
