8.00x1019 molecules HCN to grams

g HCN = MW HCN g/mole HCN * [8.00E19 molecules/6.022E23 molecules/mole HCN] = ??

EmForm = CaxPy

moles Ca = 0.66g / atwt Ca g/mole = ??

moles P = (1.00-0.66)g / atwt P g/mole = ??

x/y = moles Ca / moles P = ?? (express as ratio of two numbers; if both are not intergers multiply each by the same number that make them integers. This ratio gives the empirical formula.

Ca3P2

Plug and SOLVE

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() .

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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