How do you balance $C a_{3} {(P O_{4})}_{2} + H_{3} P O_{4} \to C a {(H_{2} P O_{4})}_{2}$ $Ca_3(PO_4)_2 + H_3PO_4 -> Ca(H_2PO_4)_2$ ?

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How do you balance $C a_{3} {(P O_{4})}_{2} + H_{3} P O_{4} \to C a {(H_{2} P O_{4})}_{2}$ $Ca_3(PO_4)_2 + H_3PO_4 -> Ca(H_2PO_4)_2$ ?

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Answer 1 · 2018-04-24T12:50:15

${Ca}_{3} {({PO}_{4})}_{2} + 4 H_{3} {PO}_{4}$ $"Ca"_3("PO"_4)_2+4"H"_3"PO"_4"$ $\to$ $rarr$ $3 {Ca(H}_{2} {PO}_{4})_{2}$ $3"Ca(H"_2"PO"_4)_2"$

Explanation:

Given equation:

${Ca}_{3} {({PO}_{4})}_{2} + H_{3} {PO}_{4}$ $"Ca"_3("PO"_4)_2+"H"_3"PO"_4"$ $\to$ $rarr$ ${Ca(H}_{2} {PO}_{4})_{2}$ $"Ca(H"_2"PO"_4)_2"$

There are three Ca atoms on the left-hand side (LHS), and one on the right-hand side (RHS). Place a coefficient of $3$ $3$ in front of ${Ca(H}_{2} {PO}_{4})_{2}$ $"Ca(H"_2"PO"_4)_2$ .

${Ca}_{3} {({PO}_{4})}_{2} + H_{3} {PO}_{4}$ $"Ca"_3("PO"_4)_2+"H"_3"PO"_4"$ $\to$ $rarr$ $3 {Ca(H}_{2} {PO}_{4})_{2}$ $color(teal)3"Ca(H"_2"PO"_4)_2"$

Now there are twelve H atoms on the RHS $(3 \times 2 \times 2)$ $(3xx2xx2)$ , and three on the LHS. Place a coefficient of $4$ $4$ in front of $H_{3} {PO}_{4}$ $"H"_3"PO"_4"$ .

${Ca}_{3} {({PO}_{4})}_{2} + 4 H_{3} {PO}_{4}$ $"Ca"_3("PO"_4)_2+color(magenta)4"H"_3"PO"_4"$ $\to$ $rarr$ $3 {Ca(H}_{2} {PO}_{4})_{2}$ $color(teal)3"Ca(H"_2"PO"_4)_2"$

Now lets count the number of atoms of each element on each side of the equation.

$LHS:$ $"LHS:"$ $3 Ca atoms$ $"3 Ca atoms"$ , $6 P atoms$ $"6 P atoms"$ , $12 H atoms$ $"12 H atoms"$ , $24 O atoms$ $"24 O atoms"$

$RHS:$ $"RHS:"$ $3 Ca atoms$ $"3 Ca atoms"$ , $6 P atoms$ $"6 P atoms"$ , $12 H atoms$ $"12 H atoms"$ , $24 O atoms$ $"24 O atoms"$

Since there are the same number of atoms of each element on both sides of the equation, it is balanced.

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How do you balance $C a_{3} {(P O_{4})}_{2} + H_{3} P O_{4} \to C a {(H_{2} P O_{4})}_{2}$ $Ca_3(PO_4)_2 + H_3PO_4 -> Ca(H_2PO_4)_2$ ?

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How do you balance Ca_3(PO_4)_2 + H_3PO_4 -> Ca(H_2PO_4)_2Ca3(PO4)2+H3PO4→Ca(H2PO4)2Ca_3(PO_4)_2 + H_3PO_4 -> Ca(H_2PO_4)_2?

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How do you balance $C a_{3} {(P O_{4})}_{2} + H_{3} P O_{4} \to C a {(H_{2} P O_{4})}_{2}$ $Ca_3(PO_4)_2 + H_3PO_4 -> Ca(H_2PO_4)_2$ ?