According to the following balanced reaction, how many moles of hno3 are formed from 8.44 moles of no2 if there is plenty of water present? 3 no2(g) + h2o(l) → 2 hno3(aq) + no(g)

Considering the definition of reaction stoichiometry, 5.63 moles of HNO₃ are formed from 8.44 moles of NO₂ if there is plenty of water present.

The balanced reaction is:

3 NO₂ + H₂O → 2 HNO₃ + NO

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

So, you can apply the following rule of three: if by reaction stoichiometry 3 moles of NO₂ produce 2 moles of HNO₃, 8.44 moles of NO₂ produces how many moles of HNO₃?

amount of moles of HNO₃= 5.63 moles

Finally, 5.63 moles of HNO₃ are formed from 8.44 moles of NO₂ if there is plenty of water present.

Learn more:

5.63 mol.

Explanation:

3NO₂(s) + H₂O(l) → 2HNO₃(aq) + NO(g),

It is clear that 3 mol of NO₂ reacts with 1 mol of H₂O to produce 2 mol of HNO₃ and 1 mol of NO.

Water is present as an excess reactant and NO₂ is limiting reactant.

3 mol of NO₂ produces → 2 mol of HNO₃, from stichiometry.

8.44 mol of NO₂ produces → ??? mol of HNO₃.

∴ The no. of moles of HNO₃ are formed = (8.44 mol)(2 mol)/(3 mol) = 5.63 mol.

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.  

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole