Question

Given the following reaction between carbonic acid and lithium hydroxide H2CO3 + LiOH → Li2CO3 + H2O a) Balance the chemical equation (2 points) b) If 10.05 mL of 3.12 M of H2CO3 required to react with 2.29 M of LiOH solution, calculate the volume of LiOH required to react with carbonic acid.

Answer 1

To balance the chemical equation H2CO3 + LiOH → Li2CO3 + H2O, follow these steps: 1. Balance the carbon atoms, then the hydrogen atoms, and finally the oxygen atoms. 2. Calculate the moles of H2CO3 and LiOH using the given concentrations and volumes. 3. Use the balanced equation to determine the stoichiometry between H2CO3 and LiOH, and calculate the volume of LiOH required to react with H2CO3.

Step-by-step explanation:

To balance the chemical equation H2CO3 + LiOH → Li2CO3 + H2O, we need to ensure that the number of atoms of each element is the same on both sides of the equation.

Here's how we can balance the equation step by step:

1. Start by balancing the carbon atoms:

Answer 2

a) To balance the chemical equation H2CO3 + LiOH → Li2CO3 + H2O, we need to make sure that the number of atoms of each element is equal on both sides of the equation.

First, let's start by counting the number of atoms for each element in the equation:

- Carbon (C): 1 on the left side and 1 on the right side

- Hydrogen (H): 2 on the left side and 2 on the right side

- Oxygen (O): 5 on the left side and 5 on the right side

- Lithium (Li): 1 on the left side and 2 on the right side

To balance the number of lithium atoms, we need to add a coefficient of 2 in front of LiOH, which gives us: H2CO3 + 2LiOH → Li2CO3 + H2O.

Now, the number of lithium atoms is balanced, but the number of hydrogen atoms is not. We have 2 hydrogen atoms on the left side and 4 on the right side. To balance the hydrogen atoms, we need to add a coefficient of 2 in front of H2O, which gives us: H2CO3 + 2LiOH → Li2CO3 + 2H2O.

Finally, all the elements are balanced, and the balanced equation is: H2CO3 + 2LiOH → Li2CO3 + 2H2O.

b) To calculate the volume of LiOH required to react with carbonic acid, we need to use the concept of molarity (M) and the equation: M1V1 = M2V2.

Given:

- Volume of H2CO3 (V1) = 10.05 mL = 10.05 x 10^(-3) L

- Molarity of H2CO3 (M1) = 3.12 M

- Molarity of LiOH (M2) = 2.29 M

Let's plug these values into the equation and solve for the volume of LiOH (V2):

M1V1 = M2V2

(3.12 M)(10.05 x 10^(-3) L) = (2.29 M)(V2)

Now, solve for V2:

V2 = (3.12 M)(10.05 x 10^(-3) L) / (2.29 M)

V2 = 13.68 x 10^(-3) L

V2 = 0.01368 L

Therefore, the volume of LiOH required to react with carbonic acid is approximately 0.01368 L or 13.68 mL.

Step-by-step explanation: