Given the heat of the following reaction: NH4Cl(aq) + NaOH(aq) ---> NH4OH(aq) +

Question

Given the heat of the following reaction:

NH4Cl(aq) + NaOH(aq) ---> NH4OH(aq) + NaCl(aq)   DeltaH = -3.6kJ

Calculate the amount of heat produced (called q in this experiment) by reacting 50.0mL of 2.0M NH4Cl with 50.0mL of 2.0M NaOH

Using the q calculated in the above question and the other information given in that problem, calculate what detlaT would be in that reaction. Assume the specific heat capacity of the solutions is 4.184J/gC.

If 10.0g of NH4NO3 is added to 50.0mL of water cause a 14.3C decrease in temperature, what is the heat of the solution for ammonium nitrate per gram? (Assume the specific heat for the solution is 4.184 J/gC)

If a 105g piece of aluminium (specific heat 0.216 cal/gC) at 25.0C absorbs 542 cal, what will its new temperature be?

Explanation / Answer

a)

millimoles of NH4cl = 50*2 = 100

millimoles of NaOH = 20*2 = 100

As you can see from the reacion

1 mole of NH4CL gives -3.6 KJ

100 millimoles gives ----------?

answer = -3.6*100*10^(-3) = -0.36 KJ

=========================================

b)

Q = msdelta T

delta T = 0.36 k /(0.1 * 53.4)(4.18)

= 16.128 degree

===========================================

c)

Q = 10 * 4.18 * (- 14.3 )

= -597.74 J

===========================

d)

delta T = 542* 4.184 /(105)(0.216)

=99.98

Tf = Ti + 99.9

=25 +99.9

=124.988 degrees

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