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3 years ago

This is the equation for the combustion of propane.

Chemistry

2 answers:

yan [13]3 years ago

8 0

Answer:

Explanation:

Reactants are the compounds which reaction produce the products. In general terms this can be expressed symbolically as follows:

reactants -> products

Other phenomena like heat are omitted because are not always present, that is, only compounds are included. Therefore, in this reaction the reactants are C3H8 (propane) and O2 ( oxygen) and the products are CO2 (carbon dioxide) and H2O (water)

Taya2010 [7]3 years ago

7 0

Answer:

$\huge \boxed{\mathrm{Option \ B}}$

$\rule[225]{225}{2}$

Explanation:

$\sf C_3 H_8 +O_2 \Rightarrow CO_2 + H_2 O$

Balancing Carbon atoms on the right side,

$\sf C_3 H_8 +O_2 \Rightarrow 3CO_2 + H_2 O$

Balancing Hydrogen atoms on the right side,

$\sf C_3 H_8 +O_2 \Rightarrow 3CO_2 +4 H_2 O$

Balancing Oxygen atoms on the left side,

$\sf C_3 H_8 +5O_2 \Rightarrow 3CO_2 +4 H_2 O$

The reactants are on the left side of the reaction:

$\sf \boxed{\sf C_3 H_8 +O_2} \Rightarrow CO_2 + H_2 O$

Propane and oxygen are the reactants.

The products are on the right side of the reaction:

$\sf C_3 H_8 +O_2 \Rightarrow \boxed{\sf CO_2 + H_2 O}$

Carbon dioxide and water are the products.

$\rule[225]{225}{2}$

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The atomic # and the mass #.

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3 years ago

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Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g

ra1l [238]

Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

$6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ$

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g) ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

$10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ$

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4 years ago

Answer both 8 and 9 will give u brain list with explanation as well so don’t just answer

goldfiish [28.3K]

Answer:

8. the answer is B.

9. the answer is A.

Explanation:

Hello!

8. In this case, by bearing to mind that the limiting reactant is always completely consumed and the excess one remain as a leftover at the end of the reaction, we can also infer that as all the limiting reactant is consumed, it must determine the maximum amount of product as the excess reactant will hypothetically produce a greater mass than expected; thus, the answer to this question is B.

9. In this case, since the mole ratio of oxygen to water is 1:2, the following proportional factor is used to calculate the produced mass of water:

$3molO_2*\frac{2molH_2O}{1molO_2}=6molH_2O$

Thus, the answer is this case is A.

Best regards!

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3 years ago

What are the three physical properties of gases

daser333 [38]

They are described through the use of four physical properties or macroscopic characteristics: pressure, volume, number of particles (chemists group them by moles) and temperature.

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