Chemistry, 22.04.2020 01:32 kendelllh

The following reaction is performed: 3H₂ + N₂ → 2NH₃ . You begin this reaction with 10.00 grams of each reactant. What is the theoretical mass of ammonia (NH₃) that can be formed

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Answer from: jace9926

The theoretical mass of ammonia is 56.6 grams when reaction is begun with 10.00 grams of each reactant.

Explanation:

Data given:

mass of reactants = 10 gram

The reactants are hydrogen and nitrogen gas.

Theoretical mass of ammonia =?

balanced chemical reaction is :

3 $H_{2}$ + $N_{2}$ ⇒ 2 N $H_{3}$

Atomic mass of hydrogen gas = 2 grams/mole

atomic mass of nitrogen gas = 28 grams/mole

number of moles of each reactant is determined:

number of moles = $\frac{mass}{atomic mass of 1 mole}$

putting the values for hydrogen gas

number of moles = $\frac{10}{2}$

number of moles = 5

number of moles of nitrogen gas = $\frac{10}{28}$

= 0.35 moles

from the balanced equation,

3 moles of hydrogen gas reacted to give 2 moles of ammonia

5 moles of hydrogen will give x moles

$\frac{2}{3}$ = $\frac{x}{5}$

3x =10

x = 3.33 moles

mass of ammonia produced = number of moles x atomic mass

= 3.33 x 17

= 56.61 grams of ammonia produced from hydrogen

For nitrogen gas,

1 mole nitrogen gas reacted to give 2 moles of ammonia gas

0.35 moles will give x moles

$\frac{2}{1} =\frac{x}{0.35}$

x = 0.7 moles of ammonia formed

mass = number of moles x atomic mass

= 0.7 x 17

= 11.9 grams

Thus the limiting reagent is hydrogen gas in the reaction.

The mass from limiting reagent is called theoretical yield = 56.61 grams

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

Yes, endothermic means it turns cold. and some chemical changes include change of tempature.

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