Answer:
a) volume of ammonium iodide required =349 mL
b) the moles of lead iodide formed = 0.0436 mol
Explanation:
The reaction is:
It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.
Let us calculate the moles of lead nitrate taken in the solution.
Moles=molarityX volume (L)
Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol
the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol
The volume of ammonium iodide required will be:
the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol
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Answer:
0.156mol
Explanation:
Number of moles of a substance can be calculated from its mass by dividing its mass by molar mass i.e.
Number of moles (n) = mass/molar mass
Molar mass of PbCl4 is as follows, where Pb = 207.2g/mol, Cl = 35.5g/lol
PbCl4 = 207.2 + 35.5(4)
= 207.2 + 142
= 349.2g/mol
Using: mole = mass/molar mass
mole = 54.32 grams ÷ 349.2g/mol
mole = 0.1555
mole = 0.156mol
