When baking soda reacts with sulfuric acid, the following reaction occurs; balance the chemical equation using the smallest possible whole numbers

When baking soda reacts with sulfuric acid, the following reaction occurs; balance the chemical equation using the smallest possible whole numbers. Answer 1 Question 2 NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + Answer 2 Question 2 H2O(l) + Answer 3 Question 2 CO2(g) In an experiment, 22.4 mL of CO2 gas is collected over a 26.0 cm column of water at 298K, where the prevailing barometric pressure is 755.2 mm Hg. The vapor pressure of water at 298K is 23.8 Torr. Calculate a. Pressure of the water column: Answer 4 Question 2 Torr [three significant digits] b. Total pressure of the CO2 and water vapor in the gas collected: Answer 5 Question 2 Torr [three significant digits] c. Partial pressure of CO2 in the gas collected: Answer 6 Question 2 atm [three significant digits] d. Moles of CO2 collected: Answer 7 Question 2 mol [three significant digits] e. Moles of NaHCO3(s) that reacted: Answer 8 Question 2 mol [three significant digits] energy of a photon

The Correct Answer and Explanation is:

Balanced Chemical Equation:

The reaction between sodium bicarbonate (baking soda) and sulfuric acid is as follows:

Unbalanced Equation: NaHCO3(s)+H2SO4(aq)→Na2SO4(aq)+H2O(l)+CO2(g)\text{NaHCO}_3(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)NaHCO3(s)+H2SO4(aq)→Na2SO4(aq)+H2O(l)+CO2(g)

To balance this equation, we ensure that the number of atoms for each element on both sides is the same.

Balance Na (Sodium): On the left, there is 1 Na from NaHCO₃. On the right, Na₂SO₄ has 2 Na, so we need 2 NaHCO₃.
Balance SO₄ (Sulfate): On the left, there is 1 SO₄ from H₂SO₄. On the right, there is 1 SO₄ from Na₂SO₄, so it is already balanced.
Balance H (Hydrogen): On the left, there are 2 H from H₂SO₄. On the right, there are 2 H from H₂O, so it’s balanced.
Balance C (Carbon): On the left, there is 1 C from NaHCO₃. On the right, there is 1 C in CO₂, so it’s balanced.
Balance O (Oxygen): On the left, there are 3 O from NaHCO₃ and 4 O from H₂SO₄. On the right, there are 4 O from Na₂SO₄, 1 O from H₂O, and 2 O from CO₂. The oxygen atoms on both sides match.

Thus, the balanced equation is: 2NaHCO3(s)+H2SO4(aq)→Na2SO4(aq)+2H2O(l)+2CO2(g)2 \text{NaHCO}_3(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2 \text{H}_2\text{O}(l) + 2 \text{CO}_2(g)2NaHCO3(s)+H2SO4(aq)→Na2SO4(aq)+2H2O(l)+2CO2(g)

Gas Collection and Pressure Calculation:

Given data:

Volume of CO₂: 22.4 mL = 0.0224 L
Height of water column: 26.0 cm
Temperature: 298K
Barometric pressure: 755.2 mm Hg
Vapor pressure of water at 298K: 23.8 Torr

(a) Pressure of the Water Column:

The pressure exerted by the water column is equal to the height of the water column, in cm, converted to Torr (since 1 cm of water is approximately equal to 0.7355 Torr). Pressure of water column=26.0 cm×0.7355 Torrcm=19.1 Torr\text{Pressure of water column} = 26.0 \, \text{cm} \times 0.7355 \, \frac{\text{Torr}}{\text{cm}} = 19.1 \, \text{Torr}Pressure of water column=26.0cm×0.7355cmTorr=19.1Torr

(b) Total Pressure of CO₂ and Water Vapor:

The total pressure of the CO₂ and water vapor is the barometric pressure minus the vapor pressure of water (since the vapor pressure contributes to the total pressure). Total pressure=755.2 mm Hg−23.8 Torr=731.4 Torr\text{Total pressure} = 755.2 \, \text{mm Hg} – 23.8 \, \text{Torr} = 731.4 \, \text{Torr}Total pressure=755.2mm Hg−23.8Torr=731.4Torr

(c) Partial Pressure of CO₂ in the Gas Collected:

The partial pressure of CO₂ is the total pressure minus the pressure exerted by the water vapor. Partial pressure of CO₂=731.4 Torr−19.1 Torr=712.3 Torr\text{Partial pressure of CO₂} = 731.4 \, \text{Torr} – 19.1 \, \text{Torr} = 712.3 \, \text{Torr}Partial pressure of CO₂=731.4Torr−19.1Torr=712.3Torr

Convert this to atmospheres: Partial pressure of CO₂=712.3760=0.937 atm\text{Partial pressure of CO₂} = \frac{712.3}{760} = 0.937 \, \text{atm}Partial pressure of CO₂=760712.3=0.937atm

(d) Moles of CO₂ Collected:

Use the ideal gas law to calculate the moles of CO₂. The ideal gas law is: PV=nRTPV = nRTPV=nRT

Where:

PPP = 0.937 atm (partial pressure of CO₂)
VVV = 0.0224 L (volume of CO₂)
RRR = 0.0821 L·atm/(mol·K) (ideal gas constant)
TTT = 298 K (temperature)

Rearranging the ideal gas law to solve for nnn: n=PVRT=(0.937 atm)(0.0224 L)(0.0821 L\cdotpatm/mol\cdotpK)(298 K)=0.001 moln = \frac{PV}{RT} = \frac{(0.937 \, \text{atm})(0.0224 \, \text{L})}{(0.0821 \, \text{L·atm/mol·K})(298 \, \text{K})} = 0.001 \, \text{mol}n=RTPV=(0.0821L\cdotpatm/mol\cdotpK)(298K)(0.937atm)(0.0224L)=0.001mol

(e) Moles of NaHCO₃(s) that Reacted:

From the balanced equation, we see that 2 moles of NaHCO₃ react with 2 moles of CO₂. So, the moles of NaHCO₃ that reacted is the same as the moles of CO₂ collected. Moles of NaHCO₃=0.001 mol\text{Moles of NaHCO₃} = 0.001 \, \text{mol}Moles of NaHCO₃=0.001mol

Energy of a Photon

The energy of a photon is given by the equation: E=hνE = h \nuE=hν