(NH4)3PO4 + Pb(NO3)4 Pb3(PO4)4 + NH4NO3
A. What type of chemical reaction is this?
B. How many grams of ammonium nitrate can be produced if 393.6 g of ammonium phosphate are reacted with excess lead (II) nitrate? (5 points) ANSWER
C. How many grams of lead (II) nitrate would be needed to react with excess ammonium phosphate to make 247.3g of lead (II) phosphate? (5 points) ANSWER
The correct answer and explanation is :
Let’s break this down step by step:
A. Type of Chemical Reaction
This reaction is a double replacement (or double displacement) reaction.
In this reaction, two ionic compounds exchange ions to form two new compounds. Here, ammonium phosphate reacts with lead(IV) nitrate to form lead(IV) phosphate and ammonium nitrate.
Balanced Equation:
$$
4 (NH_4)_3PO_4 + 3 Pb(NO_3)_4 \rightarrow Pb_3(PO_4)_4 + 12 NH_4NO_3
$$
B. Grams of Ammonium Nitrate (NH₄NO₃) produced from 393.6 g of Ammonium Phosphate
- Molar Masses:
- $(NH_4)_3PO_4$:
$(3 \times 18.04) + 30.97 + (4 \times 16.00) = 149.09 \, \text{g/mol}$ - $NH_4NO_3$:
$18.04 + 62.00 = 80.04 \, \text{g/mol}$
- Moles of Ammonium Phosphate:
$$
\frac{393.6 \, \text{g}}{149.09 \, \text{g/mol}} = 2.64 \, \text{mol}
$$
- Mole Ratio (from balanced equation):
$$
4 \text{ mol } (NH_4)_3PO_4 : 12 \text{ mol } NH_4NO_3 \Rightarrow 1 : 3
$$
So,
$$
2.64 \text{ mol } \times 3 = 7.92 \text{ mol } NH_4NO_3
$$
- Mass of NH₄NO₃ Produced:
$$
7.92 \, \text{mol} \times 80.04 \, \text{g/mol} = \boxed{633.9 \, \text{g}}
$$
C. Grams of Lead(IV) Nitrate (Pb(NO₃)₄) needed to produce 247.3 g of Lead(IV) Phosphate (Pb₃(PO₄)₄)
- Molar Masses:
- $Pb_3(PO_4)_4$:
$(3 \times 207.2) + (4 \times 94.97) = 1001.5 \, \text{g/mol}$ - $Pb(NO_3)_4$:
$207.2 + (4 \times 62.00) = 455.2 \, \text{g/mol}$
- Moles of Lead(IV) Phosphate:
$$
\frac{247.3 \, \text{g}}{1001.5 \, \text{g/mol}} = 0.247 \, \text{mol}
$$
- Mole Ratio (from balanced equation):
$$
3 \text{ mol } Pb(NO_3)_4 : 1 \text{ mol } Pb_3(PO_4)_4
$$
So,
$$
0.247 \text{ mol } \times 3 = 0.741 \text{ mol } Pb(NO_3)_4
$$
- Mass of Pb(NO₃)₄ Needed:
$$
0.741 \, \text{mol} \times 455.2 \, \text{g/mol} = \boxed{337.3 \, \text{g}}
$$
Summary:
- A. Double replacement reaction
- B. 633.9 g of NH₄NO₃ can be produced.
- C. 337.3 g of Pb(NO₃)₄ are needed.
Explanation (300 Words)
The given reaction between ammonium phosphate $(NH_4)_3PO_4$ and lead(IV) nitrate $Pb(NO_3)_4$ is a classic double replacement reaction, where the cations and anions of two reactants switch places to form new compounds. In this case, lead(IV) phosphate $Pb_3(PO_4)_4$ and ammonium nitrate $NH_4NO_3$ are formed. This reaction also involves the formation of a precipitate—lead(IV) phosphate is insoluble in water, making it a precipitation reaction as well.
To calculate the amount of ammonium nitrate produced, we first find the number of moles of ammonium phosphate from the given mass and its molar mass. Using the balanced equation, we use the mole ratio (4:12 or 1:3) to find the moles of ammonium nitrate, then convert that to grams using its molar mass. This results in 633.9 grams of ammonium nitrate.
For the second part, we determine how much lead(IV) nitrate is required to produce 247.3 grams of lead(IV) phosphate. First, we calculate the moles of lead phosphate, then apply the mole ratio (3:1) from the balanced equation to find moles of lead nitrate needed. Finally, we multiply by the molar mass of lead nitrate to find the required mass: 337.3 grams.
Stoichiometry helps us understand and predict chemical quantities in reactions, ensuring correct proportions are used and waste is minimized—vital in both lab and industrial chemistry.