calculations? (1 Viewer)

[Mr Chi]

1) calculate the theoretical mass of ethanol required to heat 200ml of water from 21.0 degrees c to 45.0 degrees c. (ethanol 1367 KJ/ mole)

2) Calculate the volume of gas produced at 25 degrees c and 100kpa by the reaction of 0.56 g of zinc with 200 ml of hydrochloric acid.

if anyone could help me, much appreciated thanks

 

[Dreamerish*~]

1. ΔH = MCΔT

ΔH = 200 x 4.18 x 24

= 20064 J

= 20.064 kJ

20.064/1367 = 0.0147

. : theoretically 0.0147 moles of ethanol is needed.

0.0147 x (12 x 2 + 1.008 x 6 + 16) = 0.68 g (2 d.p.)

. : theoretically 0.68 g of ethanol is needed.


2. Zn + 2HCl → ZnCl₂ + H₂

Does the HCl come with a concentration? I'm assuming it isn't pure HCl. It usually isn't. Anyhow, assuming that all of the 0.56 g of zinc reacts:

n = m/M

n of Zinc = 0.56/65.39

= 0.0086

The molar ratio of Zn to H₂ is 1:1. Therefore 0.0086 moles of H₂ is released.

At 25°C and 100 kPa, the volume of this many moles of H₂ gas is:

24.79 x 0.0086 = 0.21 L (2 d.p.)

Hope that helps.

 

[lol1127]

1. ΔH = MCΔT

ΔH = 200 x 4.18 x 24

= 20064 J

= 20.064 kJ

20.064/1367 = 0.0147

. : theoretically 0.0147 moles of ethanol is needed.

0.0147 x (12 x 2 + 1.008 x 6 + 16) = 0.68 g (2 d.p.)

. : theoretically 0.68 g of ethanol is needed.


2. Zn + 2HCl → ZnCl₂ + H₂

Does the HCl come with a concentration? I'm assuming it isn't pure HCl. It usually isn't. Anyhow, assuming that all of the 0.56 g of zinc reacts:

n = m/M

n of Zinc = 0.56/65.39

= 0.0086

The molar ratio of Zn to H₂ is 1:1. Therefore 0.0086 moles of H₂ is released.

At 25°C and 100 kPa, the volume of this many moles of H₂ gas is:

24.79 x 0.0086 = 0.21 L (2 d.p.)

Hope that helps.

Where did you get the 1367 from in the first question?????[/B]

 

[duck4]

1367 KJ/mol is the molar heat of combustion of ethanol. This is the amount of heat liberated when one mole of ethanol burns. So by dividing the amount of heat required to heat the water (20.064KJ) by the amount produced by burning one mole, you can work out how many moles need to be burnt.

 

[lol1127]

but where did we find out the molar heat of combustion of ethanol from?

 

[lol1127]

ohhhh right...okay i get it..we've been given it