Answer:
5 g of heptane were burned.
Explanation:
Given data:
Mass of CO₂ produced = 15.5 g
Mass of heptane burned = ?
Solution:
Balance chemical equation:
C₇H₁₆ + 11O₂ →7CO₂ + 8H₂O
First of all we will calculate the moles of CO₂ produced.
Number of moles = mass / molar mass
Molar of CO₂ = 44 g/mol
Number of moles = 15.5 g / 44 g/mol
Number of moles = 0.3523 mol
Now we will compare the moles of CO₂ and C₇H₁₆ from balance chemical equation
CO₂ : C₇H₁₆
7 : 1
0.3523 : 1/7 × 0.3523 = 0.05 mol
Mass of C₇H₁₆:
Number of moles = mass / molar mass
Mass = number of moles × molar mass
Mass = 0.05 mol / 100 g/mol
Mass = 5 g
Carbon Atoms are able to form 4 types of covalent bonds which ends up has covalent bonds
A nonmetal and a nonmetal will make molecular compounds like H2O and CO2
Answer; relative elements are most reactive elements and compounds may ignite spontaneously or explosively. They generally burn in water as well as the oxygen in the air
Explanation:
Answer:
The limiting reacting is O2
Explanation:
Step 1: data given
Number of moles O2 = 21 moles
Number of moles C6H6O = 4.0 moles
Step 2: The balanced equation
C6H6O + 7O2 → 6CO2 + 3H2O
Step 3: Calculate the limiting reactant
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
O2 is the limiting reactant. It will completely be consumed (21 moles).
C6H6O is in excess.
For 7 moles O2 we need 1 mol C6H6O
For 21 moles O2 we'll need 21/7 = 3 moles C6H6O
There will remain 4.0 - 3.0 = 1 mol C6H6O
Step 4: calculate products
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2
For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O
The limiting reacting is O2