whats the answer to this question 2 how do you balnce the equations and whats th

Question

whats the answer to this question 2 how do you balnce the equations and whats the ans for question 3

. Create a clear, well-labelled table to organize collected over the course of the experimentnd resenta and present the n you 2. The following three equations represent the chemical amount of final precipitate to the amount of P2Os in the sample dissolving the fertilizer and then forming the final precieadlions involved n equations so you know the mole relationships necessary t Balance tthe original fertilizer P2Os (s) H2O() 2tsPO(aq) H3PO4 (aq) + H2O(1) HPO (aq) NH3(aq) + HPOa2_ (aq15H2O(l) MgNMPO Mg2+(aq) + Use the final mass of the precipitate obtained experimentally, the balanced chemical equations, and the appropriate molar masses to determine the mass of phosphorus as P2Os in the original sample of plant food. 3.

Explanation / Answer

Ans 2 and 3

Balanced equations

P2O5(s) + 3 H2O(l) = 2 H3PO4(aq)

H3PO4 (aq)+ H2O(l) = HPO4 2-(aq)

Mg2+(aq) + NH3(aq) + HPO4 2-(aq) + 6H2O(l) = MgNH4PO4.6H2O(s)

n (MgNH4PO4.6H2O) = n (PO4 3-)

moles of (MgNH4PO4.6H2O) = moles as the original phosphate (PO4^3-) in the sample

molecular weight of MgNH4PO4.6H2O = 245.3 g/mol

Let x = mass of MgNH4PO4.6H2O

Moles of MgNH4PO4.6H2O = mass/molecular weight

= x/245.3

molar ratio for MgNH4PO4.6H2O to H3PO4 = 1: 1

Moles of H3PO4 = x/245.3

mole of P2O5 = 1/2 mol of (H3PO4)

Moles of P2O5 = 0.5x/245.3

molecular weight of P2O5 = 142 g/mol

Mass of P2O5 = moles x molecular weight

= (0.5x/245.3) * 142 = 0.289x

Let initial mass of fertilizer = y g

% of P2O5 in the fertilizer = (0.289x/y)*100%

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