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-BARSIC- [3]
3 years ago
11

Ammonium phosphate ((NH4)3 PO4) is an important ingredient to many fertilizers. It can be made by reacting phosphoric acid (H3 P

O4) with ammonia (NH3)
What mass of ammonia phosphate is produced by the reaction of 5.24 g of ammonia?
Chemistry
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

15.35 g of (NH₄)₃PO₄

Explanation:

First we need to look at the chemical reaction:

3 NH₃ + H₃PO₄ → (NH₄)₃PO₄

Now we calculate the number of moles of ammonia (NH₃):

number of moles = mass / molecular wight

number of moles = 5.24 / 17 = 0.308 moles of NH₃

Now from the chemical reaction we devise the following reasoning:

if         3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄

then   0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄

X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄

mass = number of moles × molecular wight

mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄

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Consider the following intermediate chemical equations.
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Answer: 250 kJ

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P_4(s)+6Cl_2\rightarrow 4PCl_3  \Delta H_1=-2439kJ (1)

4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)  \Delta H_2=3438kJ (2)

Net chemical equation:

PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)  \Delta H=? (3)

Adding 1 and 2 we get,

4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2 \Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ   (4)

Now dividing equation (4) by 4, we get

PCl_5(g)\rightarrow PCl_3(g)+Cl_2

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8 0
2 years ago
Despite the superior intellect of our lab workers, none of the problems above actually caused the discrepancies. This occurs som
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Haven't evaporated all of the water

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6 0
3 years ago
What is the freezing point of a solution of 465 g of sucrose c12h22o11 dissolved in 575 ml of water?
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Answer:

The freezing point of the solution is - 4.39 °C.

Explanation:

We can solve this problem using the relation:

<em>ΔTf = (Kf)(m),</em>

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

density of water = 1 g/mL.

<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>

m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.

<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>

<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>

<em>∴ The freezing point of the solution is - 4.39 °C.</em>

6 0
2 years ago
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