Answer to Question #141573 in General Chemistry for calista

Question

10g of sodium bicarbonate reacts with 1.0L of 0.25 mol/L'Sulfuric acid at STP conditions what is the volume of carbon dioxide produced

Answer 1

Q141573

10g of sodium bicarbonate reacts with 1.0L of 0.25 mol/L Sulphuric acid at STP conditions what is the volume of carbon dioxide produced.

Solution:

Step 1: Balancing the Chemical equation.

Our first step will be to write a balanced equation between sodium bicarbonate and sulfuric acid.

H₂SO₄ is sulfuric acid. When we add acid to a compound containing carbonate ion then carbon dioxide (CO₂) and water molecule (H₂O) is given out as a product.

NaHCO₃ (aq) + H₂SO₄ (aq) ----> Na₂SO₄ (aq) + CO₂ (g) + H₂O (l)

Na⁺ replaces H⁺ in H₂SO₄ and forms Na₂SO_4..

H⁺ from acid combines with HCO₃ ^- and forms H₂O and CO₂ .

Now balance the given chemical equation.

There is 1 Na on left side and 2 Na on right side. For balancing the Na on both side, we should have 2Na on left side too. We can do this by writing 2 in front of NaHCO₃ .

2NaHCO₃ (aq) + H₂SO₄ (aq) ----> Na₂SO₄ (aq) + CO₂ (g) + H₂O (l)

Next we balance C on both the side. There are 2 C on left side and 1 C on right side.

For balancing C we need to write 2 in front of CO₂ .

2NaHCO₃ (aq) + H₂SO₄ (aq) ----> Na₂SO₄ (aq) + 2CO₂ (g) + H₂O (l)

Next we balance the H on both the side. There are 4 H on left side and 2 H on right side.

We can balance the H by writing proper coefficient in front of H₂O.

If we write 2 in front of H₂O, then H is going be equal on both side.

2NaHCO₃ (aq) + H₂SO₄ (aq) ----> Na₂SO₄ (aq) + 2CO₂ (g) + 2 H₂O (l)

Step 2 : Find the moles of NaHCO₃ and H₂SO₄ from information given in the question.

We are told that 10g of sodium bicarbonate reacts with 1.0L of 0.25 mol/L of Sulfuric acid.

Molar mass of NaHCO₃

= 1 * atomic mass of Na + 1 * atomic mass of H + 1 * atomic mass of C+ 3 * atomic mass of O

= 1 * 22.99g/mol + 1 * 1.0079g/mol + 1 * 12.011g/mol + 3 * 15.999g/mol

=22.99 g/mol + 1.0079 g/mol + 12.011 g/mol + 47.997 g/mol

= 84.0059 g/mol

Using 84.0059g/mol convert 10g NaHCO₃ to moles.

Moles of NaHCO₃ = 10g NaHCO₃ * 1mol NaHCO₃ /84.0059 g NaHCO₃ = 0.1190 mol

Using formula of molarity find the moles of H₂SO₄ in 1.0L of 0.25 mol/L of H₂SO₄ .

Molarity = mol of solute /Volume in ‘L’ ;

Substitute molarity = 0.25mol/ L and volume = 1.0L in this formula we have

0.25mol/L : = moles of H₂SO₄ /1.00L

After solving this we have, moles of H₂SO₄ = 0.25 mol

Step 3: Find the limiting reactant.

We will find the moles of CO₂ from each reactant considering other reactant is in excess. The one which gives less moles of CO₂ will be the limiting reactant and moles of CO₂ formed actually will be the one obtained from limiting reactant.

2NaHCO₃ + H₂SO₄ (aq) ----> 1Na₂SO₄ (aq) + 2CO₂ (g) + 2 H₂O (l)

mole ratio of NaHCO₃ and CO₂ is 2:2 = 1:1,

So moles of CO₂ from NaHCO₃ = 0.1190 mol NaHCO₃ * 1 mol CO₂ / 1 mol NaHCO₃ = 0.1190 mol CO₂ .

mole ratio of H₂SO₄ and CO₂ in the reacton is 1 : 2

So, moles of CO₂ from H₂SO₄ = 0.25 mol H₂SO₄ * 2 mol CO₂ / 1 mol H₂SO₄ = 0.50 mol CO₂ .

We can see that mole of CO₂ calculated using NaHCO₃ is less. So NaHCO₃ is the limiting

reactant and H₂SO₄ is in excess.

So, moles of CO₂ actually formed is 0.1190 moles.

Step 4 : Using ideal gas equation, find the volume of CO₂ gas.

At STP, P = 1atm and T = 273.15K (0⁰ C)

The Ideal gas equation is,

PV = nRT; where P is pressure, V is volume,

T is temperature in Kelvin and R = 0.08206 L-atm/mol-K

Substitute, P =1atm, n = 0.1190 mol, T = 273.15K and R = 0.08206 L-atm/mol-K in ideal gas equation, we have

1atm * V = 0.1190mol * 273.15K * 0.08206 L-atm/mol-K

V = 0.1190mol * 273.15K * 0.08206 L-atm/mol-K / 1atm = 2.667 L

In question all the quantities are in 2 significant figure, so our final answer must also be in

2 significant figure.

Hence the volume of CO₂ given out at STP is 2.7L