Question

What's the mass of Mg3(PO4)2 in 0,4 mol?

Answer 1

M (Mg3(PO4)2) = (24*3)+((31+16*4)*2)) = 262 г/моль

m (Mg3(PO4)2) = 262 г/моль * 0,4 моль = 104,8 г

Answer 2

To calculate the mass of Mg3(PO4)2 in 0.4 mol, you need to first determine the molar mass of the compound and then use it to find the mass.

The molar mass of a compound is the sum of the atomic masses of all the atoms present in the chemical formula.

Mg3(PO4)2 consists of:

• 3 magnesium atoms (Mg) with a molar mass of approximately 24.31 g/mol each.
• 2 phosphate ions (PO4), where each phosphate ion consists of 1 phosphorus atom (P) with a molar mass of approximately 30.97 g/mol and 4 oxygen atoms (O) with a molar mass of approximately 16.00 g/mol each.

Now, calculate the molar mass of Mg3(PO4)2:

Molar mass of Mg3(PO4)2 = (3 * molar mass of Mg) + (2 * (molar mass of P + 4 * molar mass of O)) = (3 * 24.31 g/mol) + (2 * (30.97 g/mol + 4 * 16.00 g/mol)) = 72.93 g/mol + 2 * (30.97 g/mol + 64.00 g/mol) = 72.93 g/mol + 2 * 94.97 g/mol = 72.93 g/mol + 189.94 g/mol = 262.87 g/mol

Now that we have the molar mass of Mg3(PO4)2, we can find the mass of 0.4 mol:

Mass = Number of moles * Molar mass Mass = 0.4 mol * 262.87 g/mol Mass = 105.148 g

So, the mass of Mg3(PO4)2 in 0.4 mol is approximately 105.148 grams.