Iron is obtained from an iron ore by reduction with carbon monoxide. The overall

Question

Iron is obtained from an iron ore by reduction with carbon monoxide. The overall reaction is Fed) (s) + 3CO (g) 2Fe (s) + 3CO2 (g) Calculate the AHenn for the above reaction. Substance ARF wJimal+3(-105)a t 3(-893.5 -11Zd. S -gas.st330)-) Q + Fe O:(s) -825.5 CO(g) Fe(s) CO2 -110.5 -as -393.5 2.56 g of sulfur was burned in a bomb calorimeter with excess O (p). The t d from 21.25 °C to 26.72 °C. The bomb calorimeter has a heat capacity of 923 J/C. Calculate the amount of heat produced per mole of SO; formed for the following reaction. antty of heat (Ayam)(tempenanyJ s(s) + 02 (g)--S02 (g) (as1SN7) 14.003a 0 ml of 0.200 M RbOH (aq) is mixed with 100.0 ml of 0.400 M HBr (aq) in a coffee cug calon meter. If the temperature of each of the two solutions was 2440 before maxing, and the temperature rises to 26.18 calculate the enthalpy change (AH) for the reaction in kilojoules per m (kJ/mol) of HBr. RbOH (aq) + HBr (aq) H2O (l) + RbBr (aq) (Assume that the specific heat of the solution is 4.184 J/g. 'C, the volumes are additive, the densit same as water (1.00 g'ml), and no heat was lost to the surroundings)

Explanation / Answer

Solution:- (2) heat gained by calorimeter = heat capacity of calorimeter x change in temperature

change in temperature = 26.72 - 21.25 = 5.47 0C

heat capacity of calorimeter = 923 J/0C

heat gained by calorimeter = (923 J/0C)*5.47 0C = 5048.81 J = 5.05 kJ

2.56 g of S were used to form SO2. The given balanced equation is..

S + O2 ----> SO2

2.56 g S x (1mol S/32 g S) x (1 mol SO2/1mol S) = 0.08 mol SO2

So, delta H of the reaction = 5.05 kJ/0.08 mol = -63.125 kJ/mol

sign of delta H is negative since the reaction is exothermic.

(3) Volume of solution = 200.0 mL + 100.0 mL = 300.0 mL

density of solution = 1.00 g/mL

so, mass of solution = 300.0 mL x 1.00 g/mL = 300.0 g

delta T = 26.18 - 24.40 = 1.78 0C

q = m c delta T

so, heat gained by solution = 300.0 g x (4.184 J/g.0C) x 1.78 0C = 2234.256 J = 2.23 kJ

Moles of HBr added = 0.1000 L x 0.400 mol/L = 0.0400 mol

So, delta H for the reaction = 2.23 kJ/0.0400 mol

delta H for the reaction = -55.75 kJ/mol

sign of delta H is negative since the reaction is exothermic.

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