During an experiment, a student adds 0.601 g of calcium metal to 150.0 mL of 1.6

Question

During an experiment, a student adds 0.601 g of calcium metal to 150.0 mL of 1.65 M HCI. The student observes a temperature increase of 13.0 degree C for the solution. Assuming the solution's final volume is 150.0 mL, the density is 1.00 g/mL, and the specific heat is 4.184 J/(g- degree C), calculate the heat of the reaction, delta Hr times n. Ca(s)+ 2H+ (aq) rightarrow Ca2+(aq)+ H2(g) The neutralization of H3PO4 with KOH is exothermic. 60.0 mL of 0.207 M H3PO4 is mixed with 60.0 mL of 0.620 M KOH initially at 21.34 degree C. Predict the final temperature of the solution if its density is 1.13 g/mL and its specific heat is 3.78 J/(g. degree C) Assume that the total volume is the sum of the individual volumes.

Explanation / Answer

1) heat released = specific heat of solution x temp change x mass of solution

= 4.184 x 150 x 13   

= 8159 Joules = 8.159 KJ

Ca moles = 0.601/40.078 = 0.015 ,

dH of reaction per mole = 8.159/0.015 = 543.93 KJ/mol

2) moles of H3PO4 = 0.207 x60/1000 = 0.01242, KOH moles = 0.62 x0.06 =0.0372

now dH = 173.2 x 0.0124 = 2.1477 KJ

total vol = 120 ml , mass = 120 x 1.13 = 135.6

heat = 2.1477 KJ = 2147.7 = 135.6 x 3.78 x ( T-21.34)

T = 25.53 C

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