A 2.36-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3

In this experiment, carbon dioxide and water vapors combine to form H2CO3. After decomposition, the Na2CO3 had a mass of 1.57 grams.

Determine the mass of the H2CO3 produced.
Calculate the percentage yield of H2CO3 for the reaction. Show your work or describe the calculation process in detail.

Explanation:

The equation is

2NaHCO3 → Na2CO3 + H2CO3

2.36-gram sample of NaHCO3

moles of NaHCO3=mass/molar mass=2.36g/84.0 g/mol=2.8.01x10-2moles

mole ratio between NaHCO3 and H2CO3 is 2:1

Thus

moles of the H2CO3=1/2 x 2.81x10-2moles=1.4x10-2moles

mass=moles x molar mass

Expected/Theoretical mass=1.4x10-2moles x  62.03 g/mol=0.868g

After decomposition, the sample weighed 1.57 grams

meaning, the mass of H2CO3 is 2.36g-1.57g=0.79g---actual value

thus

percentage yield=actaul value/theoretical value x100%

=0.79/0.868x100%=91.01%

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

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