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Sagot :
Let's go through the step-by-step solution to determine which element is not conserved in the chemical equation:
[tex]\[ 3 \, \text{LiOH} + \text{H}_3\text{PO}_4 \rightarrow \text{Li}_2\text{PO}_4 + 3 \, \text{H}_2\text{O} \][/tex]
1. Count the number of each type of atom on the reactants' side:
- Lithium (Li): There are 3 moles of [tex]\(\text{LiOH}\)[/tex]. Each mole of [tex]\(\text{LiOH}\)[/tex] contains 1 [tex]\(\text{Li}\)[/tex] atom. So, the total Li atoms are [tex]\(3 \times 1 = 3 \text{ Li atoms}\)[/tex].
- Hydrogen (H): We have 3 moles of [tex]\(\text{LiOH}\)[/tex] and 1 mole of [tex]\(\text{H}_3\text{PO}_4\)[/tex]. Each mole of [tex]\(\text{LiOH}\)[/tex] has 1 [tex]\(\text{H}\)[/tex] atom, and each mole of [tex]\(\text{H}_3\text{PO}_4\)[/tex] has 3 [tex]\(\text{H}\)[/tex] atoms. So, the total H atoms are [tex]\(3 \times 1 + 1 \times 3 = 3 + 3 = 6 \text{ H atoms}\)[/tex].
- Phosphorus (P): There is 1 mole of [tex]\(\text{H}_3\text{PO}_4\)[/tex] and each mole has 1 [tex]\(\text{P}\)[/tex] atom. So, the total P atoms are [tex]\(1 \times 1 = 1 \text{ P atom}\)[/tex].
- Oxygen (O): We have 3 moles of [tex]\(\text{LiOH}\)[/tex] and 1 mole of [tex]\(\text{H}_3\text{PO}_4\)[/tex]. Each mole of [tex]\(\text{LiOH}\)[/tex] has 1 [tex]\(\text{O}\)[/tex] atom, and each mole of [tex]\(\text{H}_3\text{PO}_4\)[/tex] has 4 [tex]\(\text{O}\)[/tex] atoms. So, the total O atoms are [tex]\(3 \times 1 + 1 \times 4 = 3 + 4 = 7 \text{ O atoms}\)[/tex].
2. Count the number of each type of atom on the products' side:
- Lithium (Li): There is 1 mole of [tex]\(\text{Li}_2\text{PO}_4\)[/tex] and each mole has 2 [tex]\(\text{Li}\)[/tex] atoms. So, the total Li atoms are [tex]\(1 \times 2 = 2 \text{ Li atoms}\)[/tex].
- Hydrogen (H): There are 3 moles of [tex]\(\text{H}_2\text{O}\)[/tex] and each mole of [tex]\(\text{H}_2\text{O}\)[/tex] has 2 [tex]\(\text{H}\)[/tex] atoms. So, the total H atoms are [tex]\(3 \times 2 = 6 \text{ H atoms}\)[/tex].
- Phosphorus (P): There is 1 mole of [tex]\(\text{Li}_2\text{PO}_4\)[/tex] and each mole has 1 [tex]\(\text{P}\)[/tex] atom. So, the total P atoms are [tex]\(1 \times 1 = 1 \text{ P atom}\)[/tex].
- Oxygen (O): We have 1 mole of [tex]\(\text{Li}_2\text{PO}_4\)[/tex] and 3 moles of [tex]\(\text{H}_2\text{O}\)[/tex]. Each mole of [tex]\(\text{Li}_2\text{PO}_4\)[/tex] has 4 [tex]\(\text{O}\)[/tex] atoms, and each mole of [tex]\(\text{H}_2\text{O}\)[/tex] has 1 [tex]\(\text{O}\)[/tex] atom. So, the total O atoms are [tex]\(1 \times 4 + 3 \times 1 = 4 + 3 = 7 \text{ O atoms}\)[/tex].
3. Compare the counts of atoms on both sides:
- Lithium (Li): Reactants: 3, Products: 2
- Hydrogen (H): Reactants: 6, Products: 6
- Phosphorus (P): Reactants: 1, Products: 1
- Oxygen (O): Reactants: 7, Products: 7
We see that the number of Lithium (Li) atoms is not conserved as there are 3 Li atoms on the reactants' side but only 2 Li atoms on the products' side. Therefore, the element that is not conserved in this unbalanced equation is:
D. Li
[tex]\[ 3 \, \text{LiOH} + \text{H}_3\text{PO}_4 \rightarrow \text{Li}_2\text{PO}_4 + 3 \, \text{H}_2\text{O} \][/tex]
1. Count the number of each type of atom on the reactants' side:
- Lithium (Li): There are 3 moles of [tex]\(\text{LiOH}\)[/tex]. Each mole of [tex]\(\text{LiOH}\)[/tex] contains 1 [tex]\(\text{Li}\)[/tex] atom. So, the total Li atoms are [tex]\(3 \times 1 = 3 \text{ Li atoms}\)[/tex].
- Hydrogen (H): We have 3 moles of [tex]\(\text{LiOH}\)[/tex] and 1 mole of [tex]\(\text{H}_3\text{PO}_4\)[/tex]. Each mole of [tex]\(\text{LiOH}\)[/tex] has 1 [tex]\(\text{H}\)[/tex] atom, and each mole of [tex]\(\text{H}_3\text{PO}_4\)[/tex] has 3 [tex]\(\text{H}\)[/tex] atoms. So, the total H atoms are [tex]\(3 \times 1 + 1 \times 3 = 3 + 3 = 6 \text{ H atoms}\)[/tex].
- Phosphorus (P): There is 1 mole of [tex]\(\text{H}_3\text{PO}_4\)[/tex] and each mole has 1 [tex]\(\text{P}\)[/tex] atom. So, the total P atoms are [tex]\(1 \times 1 = 1 \text{ P atom}\)[/tex].
- Oxygen (O): We have 3 moles of [tex]\(\text{LiOH}\)[/tex] and 1 mole of [tex]\(\text{H}_3\text{PO}_4\)[/tex]. Each mole of [tex]\(\text{LiOH}\)[/tex] has 1 [tex]\(\text{O}\)[/tex] atom, and each mole of [tex]\(\text{H}_3\text{PO}_4\)[/tex] has 4 [tex]\(\text{O}\)[/tex] atoms. So, the total O atoms are [tex]\(3 \times 1 + 1 \times 4 = 3 + 4 = 7 \text{ O atoms}\)[/tex].
2. Count the number of each type of atom on the products' side:
- Lithium (Li): There is 1 mole of [tex]\(\text{Li}_2\text{PO}_4\)[/tex] and each mole has 2 [tex]\(\text{Li}\)[/tex] atoms. So, the total Li atoms are [tex]\(1 \times 2 = 2 \text{ Li atoms}\)[/tex].
- Hydrogen (H): There are 3 moles of [tex]\(\text{H}_2\text{O}\)[/tex] and each mole of [tex]\(\text{H}_2\text{O}\)[/tex] has 2 [tex]\(\text{H}\)[/tex] atoms. So, the total H atoms are [tex]\(3 \times 2 = 6 \text{ H atoms}\)[/tex].
- Phosphorus (P): There is 1 mole of [tex]\(\text{Li}_2\text{PO}_4\)[/tex] and each mole has 1 [tex]\(\text{P}\)[/tex] atom. So, the total P atoms are [tex]\(1 \times 1 = 1 \text{ P atom}\)[/tex].
- Oxygen (O): We have 1 mole of [tex]\(\text{Li}_2\text{PO}_4\)[/tex] and 3 moles of [tex]\(\text{H}_2\text{O}\)[/tex]. Each mole of [tex]\(\text{Li}_2\text{PO}_4\)[/tex] has 4 [tex]\(\text{O}\)[/tex] atoms, and each mole of [tex]\(\text{H}_2\text{O}\)[/tex] has 1 [tex]\(\text{O}\)[/tex] atom. So, the total O atoms are [tex]\(1 \times 4 + 3 \times 1 = 4 + 3 = 7 \text{ O atoms}\)[/tex].
3. Compare the counts of atoms on both sides:
- Lithium (Li): Reactants: 3, Products: 2
- Hydrogen (H): Reactants: 6, Products: 6
- Phosphorus (P): Reactants: 1, Products: 1
- Oxygen (O): Reactants: 7, Products: 7
We see that the number of Lithium (Li) atoms is not conserved as there are 3 Li atoms on the reactants' side but only 2 Li atoms on the products' side. Therefore, the element that is not conserved in this unbalanced equation is:
D. Li
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