Mason carried out an experiment with a plant food that contained 2.18 g of phosphate and obtained 56.4 g of the precipitate MgNH₄PO₄.6H₂O.
Let's consider the precipitation reaction of MgNH₄PO₄.6H₂O.
Mg²⁺(aq) + NH₄⁺(aq) + PO₄³⁻(aq) + 6 H₂O(l) ⇒ MgNH₄PO₄.6H₂O(s)
First, we will convert 2.18 g to moles of PO₄³⁻ using its molar mass (94.97 g/mol).
[tex]2.18 g \times \frac{mol}{94.97 g} = 0.0230 mol[/tex]
The molar ratio of PO₄³⁻ to MgNH₄PO₄.6H₂O is 1:1. The moles of MgNH₄PO₄.6H₂O produced from 0.0230 moles of PO₄³⁻ are:
[tex]0.0230 mol PO_4^{3-} \times \frac{1molMgNH_4PO_4.6H_2O}{1 mol PO_4^{3-} } = 0.0230 molMgNH_4PO_4.6H_2O[/tex]
Finally, we will convert 0.230 moles of MgNH₄PO₄.6H₂O to mass using its molar mass (245.41 g/mol).
[tex]0.230 mol \times \frac{245.41g}{mol} = 56.4 g[/tex]
Mason carried out an experiment with a plant food that contained 2.18 g of phosphate and obtained 56.4 g of the precipitate MgNH₄PO₄.6H₂O.
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