To determine the quantities of atoms in the other half of the balanced chemical equation for the reaction involving magnesium hydroxide (Mg(OH)₂) and phosphoric acid (H₃PO₄), we need to account for the number of each type of atom contributed by the reactants.
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Reactants:
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From 3 Mg(OH)₂:
- Magnesium (Mg): 3
- Oxygen (O): 3 × 2 = 6 (because each Mg(OH)₂ has 2 oxygen atoms)
- Hydrogen (H): 3 × 2 = 6 (because each Mg(OH)₂ has 2 hydrogen atoms)
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From 2 H₃PO₄:
- Hydrogen (H): 2 × 3 = 6
- Phosphorus (P): 2 × 1 = 2
- Oxygen (O): 2 × 4 = 8 (because each H₃PO₄ has 4 oxygen atoms)
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Total from reactants:
- Magnesium (Mg): 3
- Phosphorus (P): 2
- Oxygen (O): 6 (from Mg(OH)₂) + 8 (from H₃PO₄) = 14
- Hydrogen (H): 6 (from Mg(OH)₂) + 6 (from H₃PO₄) = 12
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Product side: From the original equation provided, we conclude that the total number of atoms must equal the product side.
Therefore, the other half of the balanced equation must have:
- 3 Mg
- 2 P
- 14 O
- 12 H
Thus, the correct answer is:
3 Mg, 2 P, 14 O, 12 H
5 answers