How do you balance this reaction: $? N a_{3} P O_{4} (a q) + ? C a C l_{2} (a q) \to ? N a C l (a q) + ? C a_{3} {(P O_{4})}_{2} (s)$ $?Na_3PO_4(aq) + ?CaCl_2(aq) -> ?NaCl(aq) + ?Ca_3(PO_4)_2(s)$ ?

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How do you balance this reaction: $? N a_{3} P O_{4} (a q) + ? C a C l_{2} (a q) \to ? N a C l (a q) + ? C a_{3} {(P O_{4})}_{2} (s)$ $?Na_3PO_4(aq) + ?CaCl_2(aq) -> ?NaCl(aq) + ?Ca_3(PO_4)_2(s)$ ?

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Answer 1 · 2017-01-02T17:58:53

As always, we must balance mass and charge............

Explanation:

$2 N a_{3} P O_{4} (a q) + 3 C a C l_{2} (a q) \to 6 N a C l (a q) + C a_{3} {(P O_{4})}_{2} (s) ⏐ ↓$ $2Na_3PO_4(aq) + 3CaCl_2(aq) rarr 6NaCl(aq) + Ca_3(PO_4)_2(s)darr$

Are charge and mass balanced here?

As a matter of fact, this is a simplified version of what occurs in solution. Phosphates in water are commonly the $biphosphate ion$ $"biphosphate ion"$ , $H P O_{4}^{2 -}$ $HPO_4^(2-)$ .

The reaction is probably better represented as:

$N a_{2} H P O_{4} (a q) + C a C l_{2} (a q) \to 2 N a C l (a q) + C a (H P O_{4}) (s) ⏐ ↓$ $Na_2HPO_4(aq) + CaCl_2(aq) rarr 2NaCl(aq) + Ca(HPO_4)(s)darr$

But you have to go with what your teacher says, not what I say.

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How do you balance this reaction: $? N a_{3} P O_{4} (a q) + ? C a C l_{2} (a q) \to ? N a C l (a q) + ? C a_{3} {(P O_{4})}_{2} (s)$ $?Na_3PO_4(aq) + ?CaCl_2(aq) -> ?NaCl(aq) + ?Ca_3(PO_4)_2(s)$ ?

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Explanation:

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How do you balance this reaction: ?Na3PO4(aq)+?CaCl2(aq)→?NaCl(aq)+?Ca3(PO4)2(s)?Na_3PO_4(aq) + ?CaCl_2(aq) -> ?NaCl(aq) + ?Ca_3(PO_4)_2(s)?

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How do you balance this reaction: $? N a_{3} P O_{4} (a q) + ? C a C l_{2} (a q) \to ? N a C l (a q) + ? C a_{3} {(P O_{4})}_{2} (s)$ $?Na_3PO_4(aq) + ?CaCl_2(aq) -> ?NaCl(aq) + ?Ca_3(PO_4)_2(s)$ ?