How would you balance the following equation: H3PO4 +Mg(OH)2 --> Mg3(PO4)2 +H20?

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How would you balance the following equation: H3PO4 +Mg(OH)2 --> Mg3(PO4)2 +H20?

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Answer 1 · 2015-11-05T23:29:34

Balanced Equation

$"2H"_3"PO"_4" + 3Mg(OH)"_2$ $rarr$ $"Mg"_3"(PO"_4)_2 + "6H"_2"O"$

Explanation:

$"H"_3"PO"_4" + Mg(OH)"_2$ $rarr$ $"Mg"_3"(PO"_4)_2 + "H"_2"O"$

Balance the magnesium first. There are three magnesium ions on the right side and only one on the left side. Add a coefficient of $3$ in front of $"Mg(OH)"_2"$ ..

$"H"_3"PO"_4" + 3Mg(OH)"_2$ $rarr$ $"Mg"_3"(PO"_4)_2 + "H"_2"O"$

Next balance the $"PO"_4"$ .

There are two $"PO"_4"$ ions on the right side and one on the left. Place a coefficient of $2$ in front of $"H"_3"PO"_4"$ .

$"2H"_3"PO"_4" + 3Mg(OH)"_2$ $rarr$ $"Mg"_3"(PO"_4)_2 + "H"_2"O"$

Next balance the $"H"$ . There are twelve hydrogens on the left side and two on the right. Place a coefficient of $6$ in front of $"H"_2"O"$ on the right side.

$"2H"_3"PO"_4" + 3Mg(OH)"_2$ $rarr$ $"Mg"_3"(PO"_4)_2 + "6H"_2"O"$

Balance the $"O"$ . There are fourteen oxygen atoms on both sides of the equation, so it is already balanced.

Balanced Equation

$"2H"_3"PO"_4" + 3Mg(OH)"_2$ $rarr$ $"Mg"_3"(PO"_4)_2 + "6H"_2"O"$

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How would you balance the following equation: H3PO4 +Mg(OH)2 --> Mg3(PO4)2 +H20?

1 Answer

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