Chemistry, 13.01.2021 09:40 WhatTheFangirl2927

How many grams of lead product would theoretically be produced between 17.0 g potassium iodide, KI, and 25.0 g of lead (II) nitrate, Pb(NO3)2?

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Answer from: jhenifelix

$m_{PbI_2}=23.6gPbI_2$

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

$2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2$

Thus, given the masses of potassium iodide (166.0028 g/mol) and lead (II) nitrate (331.2 g/mol), we compute the yielded moles of lead (II) iodide in order to identify the limiting reactant:

$n_{PbI_2}^{by\ KI}=17.0gKI*\frac{1molKI}{166.0gKI}*\frac{1molPbI_2}{2molKI} =0.0512molPbI_2\\\\n_{PbI_2}^{by\ Pb(NO_3)_2}=25gPb(NO_3)_2*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}*\frac{1molPbI_2}{1molPb(NO_3)_2} =0.0755molPbI_2$

Thus, since potassium iodide yields the fewest moles of product, we infer that 0.0512 moles of lead (II) iodide are theoretically yielded, so the associated mass is:

$m_{PbI_2}=0.0512molPbI_2*\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=23.6gPbI_2$

Best regards!

Answer from: Quest

the law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products. because atoms are only rearranged in a chemical reaction, there must be the same number of sodium atoms and chlorine atoms in both the reactants and products

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

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