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3 years ago

How many moles are present in 150g of ammonium hydrogen phosphite

Chemistry

1 answer:

trapecia [35]3 years ago

4 0

The moles that are present in 150 g of ammonium hydrogen phosphite is 1.136 moles

calculation

The formula of ammonium hydrogen phosphite = (NH4)2HPO4

moles = mass/molar mass

mass= 150 g

molar mass of (NH4)2HPO4 = [ ( 14+ 1x4)2) +1 + 31 + ( 4 x16)] =132 g/mol

moles is therefore = 150 g/ 132 g/mol =1.136 moles

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Value of Δ H ∘ rxn for the equation NH 3 ( g ) + 2 O 2 ( g ) ⟶ HNO3 ( g ) + H 2 O ( g )

koban [17]

Answer: - 894.6 kJ/mol.

Explanation:

Hess law is states that the changes in enthalpies in a chemical reactions is independent of the pathway between the initial and final states.

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NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ----------------------------------(**).

From the series of equations given;

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===> 2NO(g) + O2 (g) ------> 2NO2 (g). ∆H = -116.2 kJ/mol.---------------(2).

===> 3NO2 (g) + H2O (l) ---------> 2HNO3 (aq) + NO (g). ∆H = -137.3 kJ/mol.-------------------------------------(3).

The first thing to do is to multiply equation (2) by 3. Also, multiply equation (3) by 2. This will give us equation (4) and (5) respectively.

6NO + 3 O2 ----------------> 6NO2. ∆H= 3 × (-116.2 kJ/mol) = -348.6 kJ/mol. ------------------------------------(4).

6NO2 + 2 H2O ----------------> 4HNO3 + 2 NO. ∆H= 2 × (-137.3kj/mol) = -274.6 kJ/mol ---------------------------(5).

Next, add equations (4) and (5) to give;

4NO +3O2 +2H2O -------------> 4HNO3. ∆H = -623.2 kJ/mol. -----(6).

Add this equation to the equation (1) from above, we have;

4NH3 + 8O2 --------------> 4HNO3 + 4H2O. ∆H= -1789.2 kJ/mol. --------(7).

Then, divide the equation (7) above by 2 to give us back the equation (**).

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ∆H= -894.6 kJ/mol.

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