Answer:
1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2
1-2) 0.5 mole of CO2
2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O
2-2) 4 mol CO2
Step-by-step explanation:
Question 1
NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2
To balance the equation, count the number of atoms on both sides of the equation
(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)
Combining the pluses will give you the following
(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)
Both sides are the same, therefore the chemical equation is balanced (originally).
From the equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.
So that means 0.5 mole of NaHCO3 would produce 0.5 mole of CO2.
Question 2
C4H10 + O2 --> CO2 + H2O
Again, count the number of atoms on both sides of the equation
(4C, 10H, 2O) --> (1C, 2H, 3O) This time left does not equal right side
You now need to find factors that can make both sides equal.
C4H10 + O2 --> 4CO2 + H2O Now the C is balanced, let's recount
(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H) H&O is still not balanced
C4H10 + O2 --> 4CO2 + 5H2O Now the H is balanced, let's recount
(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H) O is still not balanced
C4H10 + (13/2)O2 --> 4CO2 + 5H2O Now the O is balanced
(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)
But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).
2C4H10 + 13O2 --> 8CO2 + 10H2O Now it's completely balanced!
(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H) Yayy! It's balanced.
Now, 2 mol C4H10 produces 8 mol CO2.
So 1 mol C4H10 produces 4 mol CO2.