Given the unbalanced equation -C_(3)H_(6)(g)+ldots O_(2)(g)arrow ldots H_(2)O(g)+ldots CO_(2)(g) when the equation is completely balanced using the smallest whole numbers, the coefficient of Oz 5 2 3 10

To balance the given chemical equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.<br /><br />The unbalanced equation is:<br />\[ C_{3}H_{6}(g) + O_{2}(g) \rightarrow H_{2}O(g) + CO_{2}(g) \]<br /><br />Let's balance it step by step:<br /><br />1. **Balance the carbon (C) atoms:**<br /> - There are 3 carbon atoms in \( C_{3}H_{6} \).<br /> - Therefore, we need 3 \( CO_{2} \) molecules to have 3 carbon atoms on the right side.<br /> \[ C_{3}H_{6}(g) + O_{2}(g) \rightarrow H_{2}O(g) + 3CO_{2}(g) \]<br /><br />2. **Balance the hydrogen (H) atoms:**<br /> - There are 6 hydrogen atoms in \( C_{3}H_{6} \).<br /> - Therefore, we need 3 \( H_{2}O \) molecules to have 6 hydrogen atoms on the right side.<br /> \[ C_{3}H_{6}(g) + O_{2}(g) \rightarrow 3H_{2}O(g) + 3CO_{2}(g) \]<br /><br />3. **Balance the oxygen (O) atoms:**<br /> - On the right side, we have \( 3 \times 2 = 6 \) oxygen atoms from \( CO_{2} \) and \( 3 \times 1 = 3 \) oxygen atoms from \( H_{2}O \), making a total of 9 oxygen atoms.<br /> - Therefore, we need \( \frac{9}{2} = 4.5 \) \( O_{2} \) molecules to provide 9 oxygen atoms on the left side.<br /> \[ C_{3}H_{6}(g) + 4.5O_{2}(g) \rightarrow 3H_{2}O(g) + 3CO_{2}(g) \]<br /><br />4. **Adjust to the smallest whole numbers:**<br /> - Multiply all coefficients by 2 to eliminate the fraction:<br /> \[ 2C_{3}H_{6}(g) + 9O_{2}(g) \rightarrow 6H_{2}O(g) + 6CO_{2}(g) \]<br /><br />Now, the equation is balanced:<br />\[ 2C_{3}H_{6}(g) + 9O_{2}(g) \rightarrow 6H_{2}O(g) + 6CO_{2}(g) \]<br /><br />The coefficient of \( O_{2} \) is 9. Therefore, the correct answer is not listed among the provided options.