Answer :
Final answer:
Following stoichiometry calculations in Chemistry, we find that there are 8.425 g of phosphorus in 35.8 g of Mg3(PO4)2.
Explanation:
To answer the question 'What mass of phosphorus is in 35.8 g of Mg3(PO4)2?', we need to perform a stoichiometry calculations that are part of Chemistry. First, we need to calculate the molar mass of Mg3(PO4)2. In Mg3(PO4)2, we have 3 atoms of Magnesium (Mg), 2 atoms of Phosphorus (P), and 8 atoms of Oxygen (O). The molar masses of Mg, P and O are approximately 24.305 g/mol, 30.974 g/mol and 15.999 g/mol respectively. Therefore, the molar mass of Mg3(PO4)2 would be 262.85 g/mol.
Next, we can find out how many moles of Mg3(PO4)2 is represented by 35.8 g. The number of moles = given mass/molar mass. So, the moles of Mg3(PO4)2 = 35.8 g / 262.85 g/mol = 0.136 mol.
Since there are 2 moles of P in one mole of Mg3(PO4)2, therefore, the moles of phosphorus in 35.8 g of Mg3(PO4)2 is 2 * 0.136 mol = 0.272 mol. Lastly, we multiply this number of moles by the molar mass of Phosphorus to get the mass of Phosphorus in 35.8 g of Mg3(PO4)2.
That will be 0.272 mol * 30.974 g/mol = 8.425 g.
Therefore, there is 8.425 g of phosphorus in 35.8 g of Mg3(PO4)2.
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