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lukranit [14]
3 years ago
11

Ammonium phosphate NH43PO4 is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric a

cid H3PO4 with liquid ammonia. Calculate the moles of ammonium phosphate produced by the reaction of 0.050mol of phosphoric acid. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Chemistry
1 answer:
Nataly_w [17]3 years ago
5 0

Answer:

0.050 mol

Explanation:

The reaction that takes place is:

  • H₃PO₄ + 3NH₃ → (NH₄)₃PO₄

Then we convert 0.050 moles of phosphoric acid into moles of ammonium phosphate, using the stoichiometric coefficients of the reaction:

  • 0.050 mol H₃PO₄ * \frac{1mol(NH_4)_3PO_4}{1molH_3PO_4} = 0.050 mol (NH₄)₃PO₄

Thus, the complete reaction of 0.050 moles of phosphoric acid would produce 0.050 moles of ammonium phosphate.

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Explanation:

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Answer:

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(b) 0.11 mol

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Explanation:

(a)

We use the equation given by ideal gas which follows:

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P = pressure of the gas = 1.00 atm

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R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

1.00 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1.00\times 6.0}{0.0821\times 298}=0.25mol

(b)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 0.296 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 200 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

0.296 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 200K\\\\n=\frac{0.296\times 6.0}{0.0821\times 200}=0.11mol

(c)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 30 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 250 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

30 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 250K\\\\n=\frac{30\times 6.0}{0.0821\times 250}=8.77mol

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