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