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2 years ago

How much heat is released when 15.7g of methane (c2h6) is combusted if the enthalpy of the reaction is - 1560.7 kj

Chemistry

1 answer:

lyudmila [28]2 years ago

5 0

- 407.4 kJ of heat is released.

Explanation:

We have to write the balanced equation as,

2 C₂H₆(g) + 7O₂ → 4CO₂ + 6H₂O

Here 2 moles of ethane reacts in this reaction.

Now we have to find out the amount of ethane reacted using its given mass and molar mass as,

2 mol C₂H₆ × 30.07 g of C₂H₆ / 1 mol C₂H₆ = 60.14 g of C₂H₆

Heat released = ΔH × given mass / 60.14

= - 1560. 7 kj ×15.7 g / 60. 14 g = -407. 4 kJ

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A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added

Ulleksa [173]

Answer: The pH change of the buffer is 0.30

Explanation:

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})$

$pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})$ .....(1)

We are given:

$pK_b$ = negative logarithm of base dissociation constant of aniline = 9.13

$[C_6H_5NH_3^+]=0.306M$

$[C_6H_5NH_2]=0.418M$

pOH = ?

Putting values in equation 1, we get:

$pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH_{initial}=14-8.99=5.01$

To calculate the molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-$

Initial: 0.418 0.124 0.306

Final: 0.294 - 0.430

Calculating the pOH by using using equation 1:

$pK_b$ = negative logarithm of base dissociation constant of aniline = 9.13

$[C_6H_5NH_3^+]=0.430M$

$[C_6H_5NH_2]=0.294M$

pOH = ?

Putting values in equation 1, we get:

$pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH_{final}=14-9.29=4.71$

Calculating the pH change of the solution:

$\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30$

Hence, the pH change of the buffer is 0.30

8 0

3 years ago

How many grams of nitric acid HNO₃, are required to neutralize (completely react with) 4.30 grams of Ca(OH)2 according to the ac

Brrunno [24]

Answer:

7.32g of HNO3 are required.

Explanation:

1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.

From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:

• starting with the 4.30 grams of Ca(OH)2.

• using the molar mass of Ca(OH)2 (74g/mol).

• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .

• using the molar mass of HNO3 (63.02g/mol).

$4.30g\text{ Ca\lparen OH\rparen}_2*\frac{1\text{ mol Ca\lparen OH\rparen}_2}{74g\text{ Ca\lparen OH\rparen}_2}*\frac{2\text{ moles HNO}_3}{1\text{ mole Ca\lparen OH\rparen}_2}*\frac{63.02g\text{ HNO}_3}{1\text{ mole HNO}_3}=7.32g\text{ HNO}_3$

So, 7.32g of HNO3 are required.

4 0

1 year ago

Please help! Question is on the bottom

choli [55]

It’s extremely bad quality I really can’t read it

8 0

3 years ago

What is the mass of 11.4 L of NO2

kodGreya [7K]

At STP volume is 22.4 L

Molar mass NO₂ = 46.0 g/mol

1 mole ---------- 22.4 L
? mole ---------- 11.4 L

moles = 11.4 * 1 / 22.4

moles = 11.4 / 22.4

= 0.5089 moles of NO₂

Mass NO₂ :

moles NO₂ * molar mass

0.5089 * 46.0

= 23.4094 g of NO₂

hope this helps!

7 0

3 years ago

Please help. Thank you guys for all your help and support. <3

miv72 [106K]

It would be “To obey the law of conservation of mass”

7 0

2 years ago

Read 2 more answers