A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the pressure is increased to 60.0 mm Hg? V1 = V2 =
P1 = P2 =
k = k =

Final volume = V₂ =  8.2L

Explanation:

Given data:

Initial volume = 12.3 L

Initial pressure = 40.0 mmHg

Final volume = ?

Final pressure = 60.0 mmHg

Solution:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

40.0 mmHg × 12.3 L = 60.0 mmHg × V₂

V₂ = 40.0 mmHg × 12.3 L / 60.0 mmHg

V₂ =  492 mmHg. L /  60.0 mmHg

V₂ =  8.2L

H2CO3 mass = 0.871 g

% yield = 91 %

Explanation:

The first step in solving the problem is to ensure the chemical equation is balanced, which it is, then solve for the mass of H2CO3 produced as follows:

2.36 g NaHCO3 x 1 mol NaHCO3/84.007g NaHCO3 x 1 mol H2CO3/2 mol NaHCO3 x 62.03 g H2CO3/1 mol H2CO3 = 0.871 g H2CO3

To calculate the percentage yield of H2CO3 for the reaction, you must use the Law of Conservation of Mass. Based on this law we know that mass is neither created nor destroyed. Given the initial mass of the two moles of the reactant NaHCO3 as 2.36g we know that the mass of the products should add up to approximately this amount, very little mass should have been lost in the form of gas. We know that 1.57 g of Na2CO3 remain after decomposition. Therefore, we can calculate the mass of H2CO3 remaining as follows:

M(reactant) = M (Na2CO3) + M (H2CO3)

2.36 g = 1.57 g + M (H2CO3)

M(reactant) - M(Na2CO3) = M (H2CO3)

2.36 g NaHCO3 - 1.57 g Na2CO3 = 0.79 g H2CO3

To calculate the percentage yield:

Percent yield = [Actual yield/Theoretical yield] x 100

Percent yield = [0.79 g/ 0.871g] x 100

Percent yield = 90.7 % = 91%