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Common Polyatomic Ions
\begin{tabular}{|c|c|c|c|}
\hline Chemical Name & Chemical Formula & Chemical Name & Chemical Formula \\
\hline acetate & [tex]$C _2 H _3 O _2^{-}$[/tex] & nitrite & [tex]$NO _2^{-}$[/tex] \\
\hline carbonate & [tex]$CO _3^{2-}$[/tex] & ammonium & [tex]$NH _4^{+}$[/tex] \\
\hline hypochlorite & [tex]$ClO^{-}$[/tex] & cyanide & [tex]$CN ^{-}$[/tex] \\
\hline hydrogen carbonate & [tex]$HCO _3^{-}$[/tex] & hydroxide & [tex]$OH^{-}$[/tex] \\
\hline
\end{tabular}

Sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. If you have 379.4 grams of calcium chloride and an excess of sodium phosphate, how much calcium phosphate can you make? Use the periodic table and the polyatomic ion resource.

A. 353.5 g
B. 379.4 g
C. 402.2 g
D. [tex]$\quad 863.4 g$[/tex]

Sagot :

GinnyAnswer
To solve the problem of how much calcium phosphate can be produced from 379.4 grams of calcium chloride when reacting with an excess of sodium phosphate, we'll follow these steps:

1. Write the balanced chemical equation for the reaction.
2. Calculate the moles of calcium chloride.
3. Determine the moles of calcium phosphate produced using stoichiometry.
4. Calculate the mass of calcium phosphate produced.

### Step 1: Balanced Chemical Equation

The balanced chemical equation for the reaction between calcium chloride (CaCl₂) and sodium phosphate (Na₃PO₄) is:

3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)₂ + 6 NaCl

This equation tells us that 3 moles of CaCl₂ react to produce 1 mole of Ca₃(PO₄)₂.

### Step 2: Determine Moles of Calcium Chloride

The molar mass of calcium chloride (CaCl₂) is 110.98 g/mol. We have 379.4 grams of CaCl₂.

Moles of CaCl₂ = (Mass of CaCl₂) / (Molar Mass of CaCl₂)
= 379.4 g / 110.98 g/mol
= 3.4186 mol

### Step 3: Determine Moles of Calcium Phosphate

From the balanced equation, 3 moles of CaCl₂ produce 1 mole of Ca₃(PO₄)₂.

Moles of Ca₃(PO₄)₂ = (Moles of CaCl₂) / 3
= 3.4186 mol / 3
= 1.1395 mol

### Step 4: Calculate Mass of Calcium Phosphate

The molar mass of calcium phosphate (Ca₃(PO₄)₂) is 310.18 g/mol.

Mass of Ca₃(PO₄)₂ = (Moles of Ca₃(PO₄)₂) × (Molar Mass of Ca₃(PO₄)₂)
= 1.1395 mol × 310.18 g/mol
= 353.5 g

So, the mass of calcium phosphate that can be produced is approximately 353.5 grams. Therefore, the correct answer is:

A. 353.5 g
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