Question

6. Ethanol, C2H5OH, is mixed with gasoline and sold as gasohol. Use the following to calculate the grams of ethanol needed to provide 358 kJ of heat. C2H&OH ()+3O2 (g)2CO2 (g) + 3H2O (g); AH = -1235 kJ 46.018 13.48 mal 9. A 3.456-g sample of a metal was heated to 105 °C in water bath and transferred to 50.2 g of water at 27.0 °C in a constant pressure calorimeter, a coffee cup calorimeter. The final temperature of the metal is 34.0 °C. Determine the specific heat capacity of the metal. meres tor wm.cai - 3.45 6 Cs (34.0c-105 c) (34.0C-24.0c) 50.22 4.18 2459.c.Cs :463 C 6.00

Answer 1

Answer 6 -

Given,

Heat required = 358 kJ

Mass of Ethanol = ?

C2H5OH (l) + 3 O2 (g)  $\rightarrow$ 2 CO2 (g) + 3 H2O (g) $\Delta$ H = -1235kJ

Heat of reation is -1235 kJ. which means 1 mole of C2H5OH (l) provide 1235 kJ of Heat.

1 kJ of Heat is provided by (1/1235) mole of Ethanol.

So, 358 kJ is provided by (1/1235) * 358 mol of Ethanol

Moles required = (1/1235) * 358 mol = 0.29 mol

We know that,

Moles = Mass / Molar Mass

Mass = Molar Mass * Moles

So, mass of Ethanol = 46.07 g/mol * 0.29 mol

Mass of Ethanol = 13.36 mol [Answer]

Answer 9 -

Given,

Mass of metal = 3.456 g

Initial temperature = 105 ^$\degree$ C

Mass of water = 50.2 g

Initial Temperature of water = 27.0 ^$\degree$ C

Final Temperature of metal = 34.0 ^$\degree$ C

Specific Heat of water = 4.186 joule/g °C

Specific Heat of metal =

The final Temperature of water will be same as Final Temperature of metal because heat released by metal is absorbed by water directly. No heat is released in the atmosphere.

We know that,

Heat = m*c*$\Delta$T

Heat = m*c*(Final T - Initial T)

where m = mass

c = specific heat

T = Temperature

As the heat released by metal is absorbed by water.

- Heat of Metal = Heat of water

Negative sign shows that heat is released by metal

- m_m*c_m*(Final T_m - Initial T_m) = m_w*c_w*(Final T_w - Initial T_w)

Put the values,

- 3.456 g *c_m*(34.0 °C - 105 °C) = 50.2 g *(4.186 joule/g °C)*(34.0 °C - 27 °C)

c_m = 50.2 g *(4.186 joule/g °C)*(34.0 °C - 27 °C)/(34.0 °C - 105 °C)*(- 3.456 g)

c_m = 5.99 joule/g °C [Answer]