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galina1969 [7]
3 years ago
7

A compound with the empirical formula NPCl2 was found in a subsequent experiment to have a molar mass of 347.64 g/mol. What is t

he molecular formula of the compound?
Chemistry
1 answer:
Irina-Kira [14]3 years ago
7 0

Answer:

                     Molecular Formula  =  N₃P₃Cl₆

Explanation:

Molecular formula is calculated by using following formula,

                    Molecular Formula  =  n × Empirical Formula  ---- (1)

Also, n is given as,

                     n  =  Molecular Weight / Empirical Formula Weight

Molecular Weight  =  347.64 g.mol⁻¹

Empirical Formula Weight  =  1 (N) + 1 (P) + 2 (Cl)  =  115.88 g.mol⁻¹

So,

                     n  =  347.64 g.mol⁻¹ ÷ 115.88 g.mol⁻¹

                     n  =  3

Putting Empirical Formula and value of "n" in equation 1,

                    Molecular Formula  = 3 × NPCl₂

                    Molecular Formula  =  N₃P₃Cl₆

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<u>Answer:</u> The enthalpy of the reaction is coming out to be -902 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

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\Delta H_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(4\times \Delta H_f_{(NO(g))})+(6\times \Delta H_f_{(H_2O(g))})]-[(4\times \Delta H_f_{(NH_3(g))})+(5\times \Delta H_f_{(O_2)})]

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\Delta H_f_{(NO(g))}=91.3kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-45.9kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(4\times (91.3))+(6\times (-241.8))]-[(4\times (-45.9))+(5\times (0))]\\\\\Delta H_{rxn}=-902kJ

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<u>Answer:</u> The amount of carbon dioxide gas produced in the reaction is 83.6 grams

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Thus, carbon is considered as a limiting reagent because it limits the formation of product.

To calculate the number of moles, we use the equation:

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So, 1.9 moles of carbon will produce = \frac{1}{1}\times 1.9=1.9moles of carbon dioxide gas

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