If 146 g of Pb(NO3)2 reacts stoichiometrically according to the balanced equation, how many grams of solid PbI2 are produced?
Pb(NO3)2(aq) + 2NaI(aq) ¡æ PbI2(s) + 2NaNO3(aq)
Molar Mass (g/mol)
Pb(NO3)2 331.21
PbI2 461.00
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821
To find the grams of solid PbI2 produced, you need to use stoichiometry and molar mass to convert the given mass of Pb(NO3)2 to the mass of PbI2.
1. First, calculate the number of moles of Pb(NO3)2 using the given mass and molar mass.
moles of Pb(NO3)2 = mass of Pb(NO3)2 / molar mass of Pb(NO3)2
2. Use the stoichiometric ratio from the balanced equation to convert moles of Pb(NO3)2 to moles of PbI2.
moles of PbI2 = moles of Pb(NO3)2 * (1 mole of PbI2 / 1 mole of Pb(NO3)2)
3. Finally, calculate the mass of PbI2 using the converted moles and molar mass of PbI2.
mass of PbI2 = moles of PbI2 * molar mass of PbI2
Let's solve it step by step:
Given:
- mass of Pb(NO3)2 = 146 g
- molar mass of Pb(NO3)2 = 331.21 g/mol
- molar mass of PbI2 = 461.00 g/mol
Step 1:
moles of Pb(NO3)2 = 146 g / 331.21 g/mol
moles of Pb(NO3)2 = 0.441 mol
Step 2:
moles of PbI2 = 0.441 mol * (1 mol of PbI2 / 1 mol of Pb(NO3)2)
moles of PbI2 = 0.441 mol
Step 3:
mass of PbI2 = 0.441 mol * 461.00 g/mol
mass of PbI2 = 203.001 g
Therefore, if 146 g of Pb(NO3)2 reacts stoichiometrically according to the given balanced equation, approximately 203.001 grams of solid PbI2 are produced.