Determining the Grams of Pb(NO34) Required to Produce 10 Grams of KNO3

In this article, we will discuss how to calculate the amount of Pb(NO34) needed to produce 10 grams of KNO3 based on the balanced chemical equation:

Pb(NO34) 4KI → PbI4 4KNO3

Step 1: Molar Mass Calculation

Firstly, we need to determine the molar masses of the compounds involved.

Molar Mass of KNO3:

K: 39.10 g/mol N: 14.01 g/mol O: 16.00 g/mol times; 3

Molar mass of KNO3 39.10 14.01 (3 times; 16.00) 39.10 14.01 48.00 101.11 g/mol

Molar Mass of Pb(NO34):

Pb: 207.2 g/mol N: 14.01 g/mol times; 4 O: 16.00 g/mol times; 12

Molar mass of Pb(NO34) 207.2 (4 times; 14.01) (12 times; 16.00) 207.2 56.04 192.00 455.24 g/mol

Step 2: Moles of KNO3 Produced

To find the number of moles of KNO3 in 10 grams:

Moles of KNO3 frac{mass}{molar ; mass} frac{10 ; g}{101.11 ; g/mol} ≈ 0.0988 mol

Step 3: Stoichiometry of the Reaction

According to the balanced equation, 4 moles of KNO3 are produced for every 1 mole of Pb(NO34). Therefore, the moles of Pb(NO34) required can be calculated as:

Moles of Pb(NO34) frac{0.0988 ; mol ; KNO_3}{4} ≈ 0.0247 mol

Step 4: Mass of Pb(NO34) Required

Finally, we can determine the mass of Pb(NO34) needed:

Mass of Pb(NO34) moles times; molar mass 0.0247 mol times; 455.24 g/mol ≈ 11.24 g

Conclusion

To produce 10 grams of KNO3, approximately 11.24 grams of Pb(NO34) are required.

Related Calculations:

Given molar mass of KNO3 101 g/mol and Pb(NO34) 455 g/mol, in 10g of KNO3:

No of KNO3 moles frac{mass}{molar ; mass} frac{10 ; g}{101 ; g/mol} ≈ 0.0988 mol

From the given equation, 4 moles of KNO3 are produced from 1 mole of Pb(NO34)

Hence 0.0988 moles of KNO3 will require Pb(NO34) of

frac{1}{4} times 0.0988 ; mol times 455 ; g/mol

0.0247 times 455 ; g ≈ 11.34 g

Note that Pb(NO34) is unstable. The most common lead nitrate is Pb(NO3)2.

While Pb(NO34) is highly unstable and not commercially available, the hypothetical reaction suggests that 1 mole of Pb(NO34) produces 4 moles of KNO3. If 10 grams of KNO3 are needed, approximately 11.34 grams of Pb(NO34) would be required.