Questions: Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients. Ca3(PO4)2(s) + SiO2(s) + C(s) -> CaSiO3(s) + P4(s) + CO(g)

Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients.

Ca3(PO4)2(s) + SiO2(s) + C(s) -> CaSiO3(s) + P4(s) + CO(g)
Transcript text: Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients. \[ \left.\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+\mathrm{CO}(g)\right] \]
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Solution

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Solution Steps

Step 1: Write the Unbalanced Equation

The given chemical equation is: \[ \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) + \mathrm{SiO}_{2}(s) + \mathrm{C}(s) \rightarrow \mathrm{CaSiO}_{3}(s) + \mathrm{P}_{4}(s) + \mathrm{CO}(g) \]

Step 2: List the Number of Atoms for Each Element
  • Reactants:

    • Ca: 3
    • P: 2
    • O: 8 (from \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)) + 2 (from \(\mathrm{SiO}_{2}\)) = 10
    • Si: 1
    • C: 1
  • Products:

    • Ca: 1
    • Si: 1
    • O: 3 (from \(\mathrm{CaSiO}_{3}\))
    • P: 4
    • C: 1
    • O: 1 (from \(\mathrm{CO}\))
Step 3: Balance the Phosphorus Atoms

To balance phosphorus, we need 4 phosphorus atoms on both sides. Since \(\mathrm{P}_{4}\) has 4 phosphorus atoms, we need 2 \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) on the reactant side: \[ 2\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) + \mathrm{SiO}_{2}(s) + \mathrm{C}(s) \rightarrow \mathrm{CaSiO}_{3}(s) + \mathrm{P}_{4}(s) + \mathrm{CO}(g) \]

Step 4: Balance the Calcium Atoms

Now, we have 6 calcium atoms on the reactant side. We need 6 \(\mathrm{CaSiO}_{3}\) on the product side: \[ 2\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) + \mathrm{SiO}_{2}(s) + \mathrm{C}(s) \rightarrow 6\mathrm{CaSiO}_{3}(s) + \mathrm{P}_{4}(s) + \mathrm{CO}(g) \]

Step 5: Balance the Silicon Atoms

We have 6 silicon atoms needed on the reactant side, so we need 6 \(\mathrm{SiO}_{2}\): \[ 2\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) + 6\mathrm{SiO}_{2}(s) + \mathrm{C}(s) \rightarrow 6\mathrm{CaSiO}_{3}(s) + \mathrm{P}_{4}(s) + \mathrm{CO}(g) \]

Step 6: Balance the Carbon and Oxygen Atoms

Now, balance the carbon and oxygen atoms. We have 16 oxygen atoms on the reactant side (16 from \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) and 12 from \(\mathrm{SiO}_{2}\)), and 18 on the product side (18 from \(\mathrm{CaSiO}_{3}\)). We need 10 \(\mathrm{CO}\) to balance the oxygen: \[ 2\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) + 6\mathrm{SiO}_{2}(s) + 10\mathrm{C}(s) \rightarrow 6\mathrm{CaSiO}_{3}(s) + \mathrm{P}_{4}(s) + 10\mathrm{CO}(g) \]

Final Answer

The balanced chemical equation is: \[ \boxed{2\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) + 6\mathrm{SiO}_{2}(s) + 10\mathrm{C}(s) \rightarrow 6\mathrm{CaSiO}_{3}(s) + \mathrm{P}_{4}(s) + 10\mathrm{CO}(g)} \]

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