Which clue can be used to identify a chemical reaction as a combustion reaction

Chemistry

2 answers:

Lelu [443]3 years ago

4 0

Answer: B- A hydrocarbon reacts with oxygen.

Explanation:

ehidna [41]3 years ago

3 0

combustion reaction are always exothermic in nature...so heat liberated is one of way to identify it

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Consider the reaction: 2 SO2(g)+O2(g)→2 SO3(g) If 285.5 mL of SO2 reacts with 158.9 mL of O2 (both measured at 315 K and 50.0 mm

Tanzania [10]

Answer:

a. Oxygen is the limiting reagent. $n_{SO_3}^{Theoretical}=8.096x10^{-4}mol SO_3$

b. $Y=58.9$ %

Explanation:

Hello,

a. Limiting reagent and sulfur trioxide's theoretical yield.

At first, we must compute the involved moles for both sulfur dioxide's and oxygen's as follows, considering the volumes in liters and the pressure in atm of 50.0mmHg*1atm/760mmHg=0.0658atm:

$n_{SO_2}=\frac{PV}{RT}=\frac{0.0658atm*0.2855L}{0.082\frac{atm*L}{mol*K}*315K} =7.273x10^{-3}molSO_2 \\n_{O_2}=\frac{PV}{RT}=\frac{0.0658atm*0.1589L}{0.082\frac{atm*L}{mol*K}*315K} =4.048x10^{-4}molO_2$

Afterwards, by considering the properly balanced chemical reaction:

$2SO_2(g)+O_2(g)-->2SO_3$

We compute the oxygen's moles that completely reacts with the previously computed $7.273x10^{-3}$ moles of $SO_2$ as follows:

$7.273x10^{-3}molSO_2*\frac{1molO_2}{2molSO_2} =3.6365x10^{-3}molO_2$

That result let us know that the oxygen is the limiting reagent since just $4.048x10^{-4}$ moles are available in comparison with the $3.6365x10^{-3}$ moles that completely would react with $7.273x10^{-3}$ moles of $SO_2$ .