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3 years ago

14

For the reaction between ammonium phosphate and lead (IV) nitrate, producing ammonium nitrate and lead (IV) phosphate, how many

moles of lead (IV) nitrate would be needed to produce 11.75 moles of ammonium nitrate

1 answer:

NemiM [27]3 years ago

3 0

Answer:

2.938 mol Pb(NO₃)₄

Explanation:

Step 1: Given data

Moles of NH₄NO₃ to be produced: 11.75 mol

Moles of Pb(NO₃)₄: ?

Step 2: Write the balanced equation

4 (NH₄)₃PO₄ + 3 Pb(NO₃)₄ = Pb₃(PO₄)₄ + 12 NH₄NO₃

Step 3: Calculate the number of moles of Pb(NO₃)₄required to produce 11.75 moles of NH₄NO₃

The molar ratio of Pb(NO₃)₄ to NH₄NO₃ is 3:12.

11.75 mol NH₄NO₃ × (3 mol Pb(NO₃)₄/12 mol NH₄NO₃) = 2.938 mol Pb(NO₃)₄

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The process involved in this problem are :

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Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

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$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

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SO;

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(b)

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(c)

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