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cluponka [151]
3 years ago
14

Write a balanced, net ionic equation for the precipitation reaction of CaCl2 and Cs3PO4 in an aqueous solution. Remember to incl

ude the proper physical states and charges of ions.
Chemistry
1 answer:
-Dominant- [34]3 years ago
5 0

Answer:

3Ca²⁺(aq) +  2PO4³⁻(aq) → Ca3(PO4)2(s)

Explanation:

The calcium of CaCl2 reacts with the phosphate ion od Cs3PO4 to produce the insoluble salt Ca3(PO4)2 and CsCl. The unbalanced reaction is:

CaCl2(aq) + Cs3PO4(aq) → Ca3(PO4)2(s) + CsCl(aq)

To balance the calciums:

3CaCl2(aq) + Cs3PO4(aq) → Ca3(PO4)2(s) + CsCl(aq)

The chlorides:

3CaCl2(aq) + Cs3PO4(aq) → Ca3(PO4)2(s) + 6CsCl(aq)

And the Cs:

3CaCl2(aq) + 2Cs3PO4(aq) → Ca3(PO4)2(s) + 6CsCl(aq)

This is the balanced reaction, the ionic equation is:

3Ca²⁺(aq) + 6Cl⁻(aq) + 6Cs⁺(aq) + 2PO4³⁻(aq) → Ca3(PO4)2(s) + 6Cs⁺(aq) + 6Cl⁻(aq)

Subtracting the ions that don't react:

<h3>3Ca²⁺(aq) +  2PO4³⁻(aq) → Ca3(PO4)2(s) </h3>

This is the net ionic equation

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What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?
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Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of Al(OH)_3 = 0.610 g

Molar mass of Al(OH)_3 = 78 g/mol

\text{Number of moles}=\frac{0.610g}{78g/mol}

Number of moles of Al(OH)_3 = 0.0078 moles

The reaction between Al(OH)_3 and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

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1 mole of  Al(OH)_3 reacts with 3 moles of HCl

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