Chemistry, 28.11.2019 19:31 gbprulesmile

Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3no2(g)+h2o(l)→2hno3(l)+no(g) suppose that 11 mol no2 and 3 mol h2o combine and react completely. how many moles of the reactant in excess are present after the reaction has completed?

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Answer from: mlittleduck6947

2 moles of the reactant are present in excess after the reaction has been completed

Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide

The equation of reaction is:

3NO₂(g) + H₂O(l) >>>>> 2HNO₃(l) + NO(g)

1 mole of H₂O reacts with 3 moles of NO₂

3 moles of H₂O will react with 9 moles of NO₂

Therefore, if 3 moles of H₂O reacts with 11 moles of NO₂( instead of 9 moles):

Number of moles of excess reactant = 11 moles - 9 moles

Number of moles of excess reactant = 2 moles

2 moles of the reactant are present in excess after the reaction has been completed

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Answer from: Marveon

2 mol NO2

Explanation:

3NO2(g)+H2O(l)→2HNO3(l)+NO(g)

from reaction 3 mol 1 mol

given 11 mol 3 mol

for 3 mol NO2 1 mol H2O

for x mol NO2 3 mol H2O

3:x = 1:3

x = 3 *3/1 = 9 mol NO2

So, for 3 mol H2O are needed only 9 mol NO2.

But we have 11 mol NO2. So, NO2 is in excess, and

11 mol NO2 - 9 mol NO2 = 2 mol NO2 will be left after reaction.

Answer from: Quest

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

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 :

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

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