What mass of ammonium phosphate can prepared from a mass of $3.49 \cdot g$ $3.49*g$ mass of phosphoric acid?

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What mass of ammonium phosphate can prepared from a mass of $3.49 \cdot g$ $3.49*g$ mass of phosphoric acid?

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Answer 1 · 2017-02-23T03:30:15

Approx. $1.2 \cdot g$ $1.2*g$ of $DAP$ $"DAP"$ ...

Explanation:

$Diammonium phosphate$ $"Diammonium phosphate"$ is the stable material (called $DAP$ $"DAP"$ ). This is the material that is used as agricultural fertilizer:

$H_{3} P O_{4} + 2 N H_{3} \to {(N H_{4})}_{2} H P O_{4}$ $H_3PO_4 + 2NH_3rarr(NH_4)_2HPO_4$

So following the given reaction, we assume that the phosphoric acid consumes a twice molar quantity of ammonia.

$Moles of phosphoric acid = \frac{3.49 \cdot g}{98.08 \cdot g \cdot m o l^{- 1}} = 0.0356 \cdot m o l$ $"Moles of phosphoric acid"=(3.49*g)/(98.08*g*mol^-1)=0.0356*mol$

And mass of ammonia consumed is a TWICE molar quantity, given the stoichiometric equation:

$0.0356 \cdot m o l \times 2 \times 17.03 \cdot g \cdot m o l^{- 1} \equiv 1.21 \cdot g$ $0.0356*molxx2xx17.03*g*mol^-1-=1.21*g$

Note the assumptions in this answer. I would consider an answer based on ${(N H_{4})}_{3} P O_{4}$ $(NH_4)_3PO_4$ to be unreasonable. You will have to see your teacher's treatment.

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What mass of ammonium phosphate can prepared from a mass of $3.49 \cdot g$ $3.49*g$ mass of phosphoric acid?

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What mass of ammonium phosphate can prepared from a mass of 3.49⋅g3.49*g mass of phosphoric acid?

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Explanation:

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What mass of ammonium phosphate can prepared from a mass of $3.49 \cdot g$ $3.49*g$ mass of phosphoric acid?