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Question
orange juice contains the triprotic acid citric acid, h₃c₆h₅o₇. petra titrates an orange juice sample with lioh. balance the equation. what is the coefficient for lithium hydroxide in the balanced molecular equation? h₃c₆h₅o₇ + ?lioh → li₃c₆h₅o₇ + __h₂o
Step1: Balance Li atoms
On the product side, $\ce{Li3C6H5O7}$ has 3 Li atoms. So, we need 3 $\ce{LiOH}$ to balance Li. Let's put 3 in front of $\ce{LiOH}$. Now the equation is: $\ce{H3C6H5O7 + 3LiOH -> Li3C6H5O7 + H2O}$
Step2: Balance H and O atoms
Left side: H from $\ce{H3C6H5O7}$ is 3 + 3 (from 3 $\ce{LiOH}$) = 6; O from $\ce{H3C6H5O7}$ is 7 + 3 (from 3 $\ce{LiOH}$) = 10.
Right side: $\ce{Li3C6H5O7}$ has H:5, O:7; $\ce{H2O}$ has H:2x, O:x (x is coefficient of $\ce{H2O}$).
Total H on right: 5 + 2x; total O:7 + x.
Set H equal: 5 + 2x = 6 → 2x = 1 → x=3? Wait, no. Wait, initial H in $\ce{H3C6H5O7}$ is 3 (from -COOH groups? Wait, formula is $\ce{H3C6H5O7}$, so total H is 3 + 5 = 8? Wait, maybe I miscalculated H. Let's re-express the formula: $\ce{H3C6H5O7}$ is $\ce{C6H8O7}$ (3 + 5 = 8 H). Then $\ce{LiOH}$: 3 molecules have 3 H. So total H on left: 8 + 3 = 11? No, wait, the reaction is acid-base neutralization. Citric acid ($\ce{H3C6H5O7}$) is triprotic, so it donates 3 H⁺, and $\ce{LiOH}$ donates 3 OH⁻, forming 3 H₂O. So the balanced equation should be:
$\ce{H3C6H5O7 + 3LiOH -> Li3C6H5O7 + 3H2O}$
Let's check H: Left: 3 (from $\ce{H3C6H5O7}$) + 3 (from 3 $\ce{LiOH}$) = 6? Wait, no, $\ce{H3C6H5O7}$ has H: 3 + 5 = 8? Wait, maybe the formula is $\ce{C6H8O7}$ (citric acid is $\ce{C6H8O7}$), so $\ce{H3C6H5O7}$ is maybe a typo, but assuming the given formula: $\ce{H3C6H5O7}$ has H: 3 + 5 = 8, O:7. $\ce{LiOH}$: 3 molecules: H:3, O:3. So total H left: 8 + 3 = 11, O:7 + 3 = 10.
Product: $\ce{Li3C6H5O7}$: H:5, O:7; $\ce{H2O}$: let's say coefficient is 3: H:6, O:3. Total H:5 + 6 = 11, O:7 + 3 = 10. Perfect! So coefficient of $\ce{H2O}$ is 3. And Li is balanced (3 in $\ce{Li3C6H5O7}$ and 3 in 3 $\ce{LiOH}$). Also, C is balanced (6 on both sides). So the balanced equation is $\ce{1H3C6H5O7 + 3LiOH -> 1Li3C6H5O7 + 3H2O}$
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