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How many mols of Mg3(PO4)2 are there in a beaker with 175.87g of Mg3(PO4)2?

User Fleurette
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Answer:

0.6689 moles of Mg3(PO4)2

Step-by-step explanation:

To calculate the number of moles of Mg3(PO4)2 in a beaker with 175.87 g of Mg3(PO4)2, we need to use the formula:

moles = mass / molar mass

The molar mass of Mg3(PO4)2 can be calculated by adding the atomic masses of all the atoms in the compound:

Mg3(PO4)2 = 3(Mg) + 2(P) + 8(O)

Mg3(PO4)2 = (3 x 24.31 g/mol) + (2 x 30.97 g/mol) + (8 x 16.00 g/mol)

Mg3(PO4)2 = 262.86 g/mol

Now we can plug in the values we have:

moles = 175.87 g / 262.86 g/mol

moles = 0.6689 mol

Therefore, there are 0.6689 moles of Mg3(PO4)2 in the beaker with 175.87 g of Mg3(PO4)2.

User Mitzie
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