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3 years ago

Propane burns in oxygen to produce carbon dioxide and steam. The unbalanced equation for this reaction is: C3H8 + O2 → CO2 + H2O

1 answer:

8 0

Here is your answer

C3H8 + O2-----> CO2 + H2O

Balancing C atoms

C3H8 + O2------> 3CO2 + H2O

Balancing H atoms

C3H8 + O2------> 3CO2 + 4H2O

Balancing O atoms

C3H8 + $\bold{5}$ O2-------> $\bold{3}$ CO2 + $\bold{4}$ H2O

Hence the equation is balanced.

HOPE IT IS USEFUL

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Geothermal energy is an alternative energy source. However, it is not resourceful enough to replace more than a minor amount of

Brums [2.3K]

A- Drilling through the Earth to reach areas is useful but expensive

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4 years ago

2CO +0, 200, what is the mole ratio of carbon monoxide to carbon

Nastasia [14]

Answer:

B. 2.2

Explanation:

6 0

3 years ago

A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon

ikadub [295]

<u>Answer:</u> The enthalpy of the reaction is 269.4 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

$q_1=c\Delta T$

where,

q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

$\Delta T$ = change in temperature = $T_2-T_1=(53.88-24.26)^oC=29.62^oC$

Putting values in above equation, we get:

$q_1=675J/^oC\times 29.62^oC=19993.5J$

To calculate the heat absorbed by water, we use the equation:

$q_2=mc\Delta T$

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(53.88-24.26)^oC=29.62^oC$

Putting values in above equation, we get:

$q_2=925g\times 4.186J/g^oC\times 29.62^oC=114690.12J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[19993.5+114690.12]J=134683.62J=134.7kJ$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat absorbed = 134.7 kJ

n = number of moles of hydrocarbon = 0.500 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{134.7kJ}{0.500mol}=269.4kJ/mol$

Hence, the enthalpy of the reaction is 269.4 kJ/mol

6 0

3 years ago

Read 2 more answers

A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant

KiRa [710]

Answer:

$V_2=2.995L\\\\W=248.5J$

Explanation:

Hello,

In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

$P_1V_1=P_2V_2$

So we solve for $V_2$ by firstly computing the initial pressure:

$P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L} =11.98atm$

$V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L$

Finally, we can compute the work by using the following formula:

$W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J$

Best regards.

4 0

3 years ago

For the following reaction, 42.2 grams of potassium hydrogen sulfate are allowed to react with 21.4 grams of potassium hydroxide

ASHA 777 [7]

Answer:

53.99g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

KHSO4(aq) + KOH(aq) —> K2SO4(aq) + H2O(l)

Step 2:

Determination of the masses of KHSO4 and KOH that reacted and the mass of K2SO4 produced from the balanced equation.

This is illustrated below:

Molar mass of KHSO4 = 39 + 1 + 32 + (16x4) = 136g/mol

Mass of KHSO4 from the balanced equation = 1 x 136 = 136g

Molar mass of KOH = 39 + 16 + 1 = 56g/mol

Mass of KOH from the balanced equation = 1 x 56 = 56g

Molar mass of K2SO4 = (39x2) + 32 + (16x4) = 174g/mol

Mass of K2SO4 from the balanced equation = 1 x 174 = 174g.

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH to produce 174g of K2SO4

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH.

Therefore, 42.2g of KHSO4 will react with = (42.2 x 56)/136 = 17.38g of KOH.

From the above calculations, we can see that only 17.38g out of 21.4g of KOH given was needed to react completely with 42.2g of KHSO4.

Therefore, KHSO4 is the limiting reactant and KOH is the excess reactant.

Step 4:

Determination of the maximum mass of K2SO4 produced from the reaction.

In this case, the limiting reactant will be used as all of it is used up in the reaction. The limiting reactant is KHSO4 and the maximum amount of K2SO4 produced can be obtained as follow:

From the balanced equation above, 136g of KHSO4 reacted to produce 174g of K2SO4.

Therefore, 42.2g of KHSO4 will react to produce = (42.2 x 174)/136 = 53.99g of K2SO4.

Therefore, the maximum amount of K2SO4 produced is 53.99g.

8 0

3 years ago