How much heat is liberated at constant pressure if 0.500 g of calcium carbonate

Question

How much heat is liberated at constant pressure if 0.500 g of calcium carbonate reacts with 12.0 mL of 0.100 M hydrochloric acid?

CaCO3(s) + 2HCl(aq)    CaCl2(aq) + H2O(l) + CO2(g); H° = –15.2 kJ

A.-7.60 kJ

B.-0.0759 kJ

C.-1.52 kJ

D.-0.0851 kJ

E.-0.00912 kJ

The reaction of iron with hydrochloric acid is represented by the following thermochemical equation.

Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g); H° = –87.9 kJ

If, in a particular experiment, 5.10 kJ of heat was released at constant pressure, what volume of H2(g), measured at STP, was produced? (R = 0.0821 L • atm/(K • mol))

A.4.22 × 102 L

B.3.86 × 102 L

C.1.42 L

D.1.30 L

E.22.4 L

How much heat is gained by zinc when 41.0 g of zinc is warmed from 29.0°C to 71.0°C? The specific heat of zinc is 0.388 J/(g · °C).

A.1.13 × 103 J

B.16.30 J

C.4.61 × 102 J

D.6.68 × 102 J

E.27.55 J

Explanation / Answer

How much heat is liberated at constant pressure if 0.500 g of calcium carbonate reacts with 12.0 mL of

0.100 M hydrochloric acid?

CaCO3(s) + 2HCl(aq)    CaCl2(aq) + H2O(l) + CO2(g); H° = –15.2 kJ

We are given equation , mass, volume and molarity of HCl

We calculate number of moles of CaCO3 and HCl. Once we get number of moles of each reactant then we calculate limiting reactant from limiting reactant.

Mol CaCO3 = Mass in g / molar mass = 0.500 g / 100.086 g permol = 0.0050 mol CaCO3

Mol HCl = Volume in L * molarity = 0.012 L * 0.100 M =0.0012 mol HCl

Lets predict limiting reactant.

We calculate moles of any product from both

Mol H2O from CaCO3

=0.0050 mol CaCO3 * 1 mol H2O / 1 mol CaCO3 = 0.0050 H2O

Mol H2O from HCl

= 0.0012 mol HCl * 1 mol H2O / 2 mol HCl = 0.0006 mol H2O

HCl gives less number of moles and so it is limiting reactant.

Now liberated heat is calculated by using moles of HCl

We know 2 mol HCl liberates = -15.2 kJ

Heat liberated by 0.0012 mol HCl

= 0.0012 mol HCl * (-15.2 kJ) /2 mol HCl

= -0.00912 kJ and the answer for it E

A.-7.60 kJ

B.-0.0759 kJ

C.-1.52 kJ

D.-0.0851 kJ

E.-0.00912 kJ

The reaction of iron with hydrochloric acid is represented by the following thermochemical equation.

Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g); H° = –87.9 kJ

If, in a particular experiment, 5.10 kJ of heat was released at constant pressure, what volume of H2(g), measured at STP, was produced? (R = 0.0821 L • atm/(K • mol))

By using delta H value we get the number of moles of H2 produced. Once we get moles of H2 then we use STP to get volume of it.

From the equation we know 1 mol of H2 produced as the expense of -87.9 kJ

Number of mole of H2 = - 5.10kJ * 1 mol H2/ - 87.9 kJ = 0.058 mol

We use ideal gas law

PV = nRT     R = 0.0821 L atm /( K .mol )

Now at STP , P = 1 atm , T = 273.15 K

Lets plug all the value in ideal gas law and get volume in L

V = nRT / P

= 0.058 mol * (0.0821L atm / K mol ) * 273.15 K / 1 atm

= 1.30 L

And the answer is D = 1.30 L

A.4.22 × 102 L

B.3.86 × 102 L

C.1.42 L

D.1.30 L

E.22.4 L

How much heat is gained by zinc when 41.0 g of zinc is warmed from 29.0°C to 71.0°C? The specific heat of zinc is 0.388 J/(g · °C).

Solution :

We use following equation to calculate heat gained by Zn

q = c m Delta T

C is specific heat of C = 0.388 J / g deg C   , Delta T = change in T = Tfinal – T initial.

m is mass in g

Lets plug all the value and find q

q = 41.0 g * 0.388 J / g deg C * (71.0 – 29.0 deg C )

= 668.136 J or 6.68 E 2 J

And hence the answer is D = 6.68 X 102 J

A.1.13 × 103 J

B.16.30 J

C.4.61 × 102 J

D.6.68 × 102 J

E.27.55 J

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