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
Solution:
The molecular formula is PbSO₄, indicating lead sulfate
Option c.
Explanation:
The percentage makeup shows that in 100 g of this compound, there are:
68.3 g of Pb, 10.6 g of S, and (100 - 68.3 - 10.6) = 21.1 g of O
To find the moles of each element, we divide by their molar masses:
68.3 g Pb / 207.2 g/mol = 0.329 moles Pb
10.6 g S / 32.06 g/mol = 0.331 moles S
21.1 g O / 16 g/mol = 1.32 moles O
Next, we find the mole ratio by dividing each by the smallest number of moles:
0.329 / 0.329 = 1 Pb
0.331 / 0.329 = 1 S
1.32 / 0.329 = 4 O
Thus, the molecular formula is PbSO₄, representing lead sulfate.
