QUESTION IMAGE
Question
write a balanced chemical equation based on the following description: propanol, c₃h₇oh(l) reacts with oxygen gas to produce carbon dioxide gas and water vapor c₃h₇oh(l) + o₂(g) →
Step1: Write the unbalanced equation
The reaction is \(C_{3}H_{7}OH(l)+O_{2}(g)\to CO_{2}(g)+H_{2}O(g)\)
Step2: Balance carbon atoms
There are 3 carbon atoms in \(C_{3}H_{7}OH\). So, we put a coefficient of 3 in front of \(CO_{2}\). The equation becomes \(C_{3}H_{7}OH(l)+O_{2}(g)\to3CO_{2}(g)+H_{2}O(g)\)
Step3: Balance hydrogen atoms
There are 8 hydrogen atoms in \(C_{3}H_{7}OH\). So, we put a coefficient of 4 in front of \(H_{2}O\). The equation is \(C_{3}H_{7}OH(l)+O_{2}(g)\to3CO_{2}(g)+4H_{2}O(g)\)
Step4: Balance oxygen atoms
On the right - hand side, the total number of oxygen atoms is \(3\times2 + 4\times1=6 + 4=10\). In \(C_{3}H_{7}OH\), there is 1 oxygen atom. So, the number of oxygen atoms from \(O_{2}\) is \(10 - 1=9\). Since each \(O_{2}\) molecule has 2 oxygen atoms, we put a coefficient of \(\frac{9}{2}\) in front of \(O_{2}\). But we usually prefer whole - number coefficients. Multiply the entire equation by 2 to get rid of the fraction.
The balanced equation is \(2C_{3}H_{7}OH(l)+9O_{2}(g)\to6CO_{2}(g)+8H_{2}O(g)\)
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\(2C_{3}H_{7}OH(l)+9O_{2}(g)\to6CO_{2}(g)+8H_{2}O(g)\)