Hydrochloric acid can be prepared using the reaction described by the chemical equation: 2 NaCl(s) + H2SO4(l) > 2 HCl(g) + Na2 SO4(s). How many grams of HCl can be prepared from 393 g of H2SO4 and 4.00 moles of NaCl?

145.84 g

Explanation:

To answer this problem first we need to determine the limiting reactant, to do so we convert 393 g of H₂SO₄ into moles using its molar mass:

4.01 moles of H₂SO₄ would react completely with (4.01 * 2) 8.02 moles of NaCl. There are not as many NaCl moles, so NaCl is the limiting reactant.

Now we calculate how many HCl moles are formed, using the number of moles of the limiting reactant and the stoichiometric coefficients of the reaction:

Finally we convert HCl moles into grams, using its molar mass:

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.

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.