What mass of Pb(NO3)2,would be required to yeild 9g of PbCl2,on d addition of excex NaCl?Pb=207,N=14,O=16,Na=23,Cl=35.5
Pb(NO3)2 + 2NaCl>>PbCl2 + 2NaNO3(aq)
so, moles Pb(NO3)2 = moles PbCl2
Moles PbCl2= 9g/(783)
MassPb(NO3)2= molesPbCl2*(molemass leadII nitrate).
To find the mass of Pb(NO3)2 required to yield 9g of PbCl2, first, we need to determine the balanced chemical equation for the reaction that occurs when Pb(NO3)2 reacts with NaCl. From the given information, we know that the elements involved in this reaction are Pb, N, O, Na, and Cl.
The balanced chemical equation for this reaction is:
2Pb(NO3)2 + 2NaCl → 2PbCl2 + 4NaNO3
According to the equation, 2 moles of Pb(NO3)2 react to produce 2 moles of PbCl2. We need to calculate the molar mass of Pb(NO3)2 to convert the mass of PbCl2 to moles.
The molar mass of Pb(NO3)2 can be calculated by summing up the atomic masses of its constituent elements:
Pb(NO3)2:
Pb: 207g/mol
N: 14g/mol
O: 16g/mol (x 3 because of 3 oxygen atoms) = 48g/mol
Molar mass of Pb(NO3)2 = 207g/mol + 48g/mol = 255g/mol
Next, we can calculate the number of moles of PbCl2 produced from the given mass of 9g using the molar mass of PbCl2:
Molar mass of PbCl2:
Pb: 207g/mol
Cl: 35.5g/mol (x 2 because of 2 chloride ions) = 71g/mol
Molar mass of PbCl2 = 207g/mol + 71g/mol = 278g/mol
Number of moles of PbCl2 = given mass / molar mass
Number of moles of PbCl2 = 9g / 278g/mol ≈ 0.0324 mol
Since the stoichiometric ratio between Pb(NO3)2 and PbCl2 is 2:2, it means that for every 2 moles of Pb(NO3)2, we get 2 moles of PbCl2.
Therefore, the number of moles of Pb(NO3)2 needed is equal to the number of moles of PbCl2:
Number of moles of Pb(NO3)2 = 0.0324 mol
Finally, we can calculate the mass of Pb(NO3)2 by multiplying the number of moles by the molar mass:
Mass of Pb(NO3)2 = number of moles * molar mass
Mass of Pb(NO3)2 = 0.0324 mol * 255g/mol ≈ 8.28g
Therefore, approximately 8.28g of Pb(NO3)2 would be required to yield 9g of PbCl2 on the addition of excess NaCl.
To find the mass of Pb(NO3)2 required to yield 9g of PbCl2, we need to consider the stoichiometry of the reaction and perform some calculations. Here's the step-by-step process:
Step 1: Write the balanced chemical equation for the reaction
Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3
Step 2: Determine the molar mass of PbCl2
The molar mass of PbCl2 is calculated by adding the atomic masses of Pb and 2 times the atomic mass of Cl:
Pb: 207 g/mol
Cl: 35.5 g/mol
Molar mass of PbCl2 = 207 + 2 * 35.5 g/mol = 278 g/mol
Step 3: Convert the given mass of PbCl2 to moles
Using the molar mass calculated in Step 2, we can convert the given mass of PbCl2 (9g) to moles:
Moles of PbCl2 = Mass / Molar mass
Moles of PbCl2 = 9g / 278 g/mol = 0.032 moles
Step 4: Use the stoichiometry to relate Pb(NO3)2 and PbCl2
From the balanced chemical equation, the stoichiometric ratio between Pb(NO3)2 and PbCl2 is 1:1. This means that 1 mole of Pb(NO3)2 produces 1 mole of PbCl2. Therefore, the moles of Pb(NO3)2 required will be the same as the moles of PbCl2:
Moles of Pb(NO3)2 = Moles of PbCl2 = 0.032 moles
Step 5: Convert the moles of Pb(NO3)2 to mass
To find the mass of Pb(NO3)2, we multiply the moles of Pb(NO3)2 by the molar mass of Pb(NO3)2:
Molar mass of Pb(NO3)2 = 207 + 2 * (14 + 3 * 16) = 207 + 2 * (14 + 48) = 207 + 2 * 62 = 207 + 124 = 331 g/mol
Mass of Pb(NO3)2 = Moles of Pb(NO3)2 * Molar mass of Pb(NO3)2
Mass of Pb(NO3)2 = 0.032 moles * 331 g/mol = 10.592 g
Therefore, 10.592 grams of Pb(NO3)2 would be required to yield 9 grams of PbCl2.