VERY URGENT! IB CHEMISTRY
0.0810 g of a group 2 metal iodide, MI2, was dissolved in water and made up to a total volume of 25.00 cm3.

Excess lead(II) nitrate solution (Pb(NO3)2(aq)) was added to the MI2 solution to form a precipitate of lead(II)

iodide (PbI2). The precipitate was dried and weighed and it was found that 0.1270 g of precipitate was obtained.

a Determine the number of moles of lead iodide formed.

b Write an equation for the reaction that occurs.

c Determine the number of moles of MI2 that reacted.

d Determine the identity of the metal, M.

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mickymike92 mickymike92 · Answer 1 · 2021-12-21T14:56:30+01:00

The number of moles of lead iodide formed is 0.000275 moles

The equation of the reaction is as follows:

MI₂ (aq) + Pb(NO₃)₂ (aq) ----> M(NO₃)₂ (aq) + PbI₂ (s)

The number of moles of MI₂ that reacted is 0.000275 moles

The group 2 metal is Calcium whose molar mass is 40.0 g

a. From the given values:

Mass of lead iodide precipitated = 0.1270 g

molar mass of lead iodide = 461 g/mol

number of moles = mass / molar mass

number of moles of lead iodide formed = 0.1270 g / 461 g /mol

number of moles of lead iodide formed = 0.000275 moles

b. The equation of the reaction shows the reactants as well as the products formed after reaction.

The general molecular equation is given as :

MI₂ (aq) + Pb(NO₃)₂ (aq) ----> M(NO₃)₂ (aq) + PbI₂ (s)

The net ionic equation is given as:

Pb²⁺ (aq) + 2 I⁻ (aq) ---> PbI₂ (s)

c. 1 mole of MI₂ reacts with 1 mole of Pb(NO₃)₂ to produce 1 mole of M(NO₃)₂ and 1 mole PbI₂

Since 0.000275 moles of PbI₂ was formed, it would require 0.000275 moles MI₂ to be formed.

Number of moles of MI₂ that reacted = 0.000275 moles

d. Mass of 0.000275 moles of MI₂ = 0.0810 g

mass of 1 mole of MI₂ = 0.0810 / 0.000275 =294.5 g

In 294.5 g of MI₂, mass composition of Iodide = 127 * 2 = 254 g

Therefore mass of the metal = mass of compound - mass of iodine in the compound

mass of metal, M = 294.5 - 254 = 40.5 g

The group 2 metal is Calcium whose molar mass is 40.0 g

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