Amount of NaHCO3 required for obtaining 0.56 litre of CO2 gas at STP

To determine the amount of NaHCO3 required to obtain 0.56 liters of CO2 gas at STP, we need to use the concept of stoichiometry and the ideal gas law.

Stoichiometry and balanced chemical equation:
The balanced chemical equation for the reaction of NaHCO3 (sodium bicarbonate) decomposing into CO2 (carbon dioxide), Na2CO3 (sodium carbonate), and H2O (water) is:

2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)

According to this equation, for every 2 moles of NaHCO3, we will obtain 1 mole of CO2 gas.

Using the ideal gas law:
The ideal gas law equation is given as:

PV = nRT

Where:
P = pressure (at STP, P = 1 atm)
V = volume of gas (0.56 liters)
n = number of moles of gas (to be determined)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (at STP, T = 273 K)

Rearranging the equation, we find:

n = PV / RT

Plugging in the given values, we have:

n = (1 atm) * (0.56 L) / (0.0821 L·atm/mol·K * 273 K)

Simplifying the expression, we get:

n ≈ 0.0236 moles

Calculating the amount of NaHCO3:
From the balanced equation, we know that 2 moles of NaHCO3 yield 1 mole of CO2. Therefore, the number of moles of NaHCO3 required can be found using the mole ratio:

0.0236 moles CO2 × (2 moles NaHCO3 / 1 mole CO2) = 0.0472 moles NaHCO3

Converting moles to grams:
To convert the moles of NaHCO3 to grams, we need to know the molar mass of NaHCO3. The molar mass of NaHCO3 is calculated by adding the atomic masses of each element:

Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol (3 oxygen atoms)

Molar mass of NaHCO3 = 22.99 + 1.01 + 12.01 + (16.00 * 3) = 84.01 g/mol

Converting moles to grams:

0.0472 moles NaHCO3 × (84.01 g NaHCO3 / 1 mole NaHCO3) ≈ 3.96 grams NaHCO3

Therefore, approximately 3.96 grams of NaHCO3 are required to obtain 0.56 liters of CO2 gas at STP.