Answer 6 -
Given,
Heat required = 358 kJ
Mass of Ethanol = ?
C2H5OH (l) + 3 O2 (g) 2 CO2 (g) + 3 H2O (g) 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 C
Mass of water = 50.2 g
Initial Temperature of water = 27.0 C
Final Temperature of metal = 34.0 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*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
- mm*cm*(Final Tm - Initial Tm) = mw*cw*(Final Tw - Initial Tw)
Put the values,
- 3.456 g *cm*(34.0 °C - 105 °C) = 50.2 g *(4.186 joule/g °C)*(34.0 °C - 27 °C)
cm = 50.2 g *(4.186 joule/g °C)*(34.0 °C - 27 °C)/(34.0 °C - 105 °C)*(- 3.456 g)
cm = 5.99 joule/g °C [Answer]
6. Ethanol, C2H5OH, is mixed with gasoline and sold as gasohol. Use the following to calculate...
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