if 5liters of hydrogen iodide gas is bubbled through 2 liters of 0.0275 M lead 2 nitrate solution at STP, how many grams of lead 2 iodide will precipitate out?
To answer this question, we need to use stoichiometry and the concept of limiting reagents.
First, let's write the balanced chemical equation for the reaction:
2HI(g) + Pb(NO3)2(aq) -> PbI2(s) + 2HNO3(aq)
From the equation, we can see that two moles of hydrogen iodide are needed to react with one mole of lead(II) nitrate to form one mole of lead(II) iodide.
Now let's calculate the moles of hydrogen iodide present. We can use the ideal gas law to convert the given volume of hydrogen iodide gas to moles.
PV = nRT
Since the gas is at STP (Standard Temperature and Pressure), we have:
P = 1 atm
V = 5 liters
R = 0.0821 L.atm/mol.K
T = 273 K
n = PV / RT
= (1 atm) * (5 L) / (0.0821 L.atm/mol.K * 273 K)
≈ 0.212 mol of HI
Now, let's calculate the moles of lead(II) nitrate present. We can use the given volume and molarity to calculate the moles.
Molarity (M) = moles / volume
0.0275 M = x moles / 2 L
x = 0.055 mol of Pb(NO3)2
Since the reaction stoichiometry is 2:1 (2 moles of HI react with 1 mole of Pb(NO3)2), the limiting reagent will be the one with the smaller number of moles.
In this case, we have 0.055 moles of Pb(NO3)2 and 0.212 moles of HI. Since we have fewer moles of Pb(NO3)2, it is the limiting reagent.
Now, let's calculate the moles of lead(II) iodide that will precipitate out. Using the stoichiometry of the balanced equation, we know that one mole of Pb(NO3)2 yields one mole of PbI2.
Therefore, the moles of PbI2 formed will be the same as the moles of Pb(NO3)2, which is 0.055 mol.
Finally, to calculate the mass of lead(II) iodide, we need to use its molar mass. From the periodic table, the molar mass of PbI2 is approximately 461 g/mol.
Mass = moles * molar mass
= 0.055 mol * 461 g/mol
≈ 25.3 grams of PbI2 will precipitate out.
So, approximately 25.3 grams of lead(II) iodide will precipitate out in the given reaction conditions.