I need a problem:

LiOH + CO2 ---> Li2CO3 + H2O

1. to balance the equation?

2. write the reasons estequiometricas with regard to Li2CO3?

3.how much do masses of Li2CO3 form if we add 37g of LiOHa to the Rxn?

thanks

What is the mass in grams of carbon dioxide that would be required to react with 87.4 g of LiOH in the following chemical reaction?

2 LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

Sure, let's go step by step to solve the problem:

1. Balancing the equation:

To balance the equation: LiOH + CO2 -> Li2CO3 + H2O

First, let's count the number of atoms on each side:

LiOH: 1 Li, 1 O, 1 H
CO2: 1 C, 2 O
Li2CO3: 2 Li, 1 C, 3 O
H2O: 2 H, 1 O

To balance the equation, let's start by balancing the carbon (C) atoms:

CO2 + LiOH -> Li2CO3 + H2O

Now let's balance the hydrogen (H) atoms:

CO2 + 2 LiOH -> Li2CO3 + H2O

Finally, let's balance the oxygen (O) atoms:

CO2 + 2 LiOH -> Li2CO3 + H2O

This is the balanced equation.

2. Stoichiometric reasons regarding Li2CO3:

In the balanced equation, the stoichiometric coefficients represent the mole ratio between the reactants and the products. From the balanced equation:

1 mole of LiOH reacts with 1 mole of CO2 to produce 1 mole of Li2CO3 and 1 mole of H2O.

This means that if you have 1 mole of LiOH and completely react it with CO2, you will produce 1 mole of Li2CO3.

3. Calculating the mass of Li2CO3 formed:

Given that you have 37g of LiOH, you can calculate the mass of Li2CO3 formed using the stoichiometric ratio from the balanced equation.

The molar mass of LiOH is approximately 23.95 g/mol for Li and 17.01 g/mol for OH. Adding them up, we get:

23.95 g/mol Li + 17.01 g/mol OH = 40.96 g/mol LiOH

To find the number of moles of Li2CO3, divide the mass of LiOH by the molar mass:

37 g LiOH / 40.96 g/mol LiOH = 0.903 mol LiOH

Using the stoichiometric ratio from the balanced equation, we know that 1 mole of LiOH produces 1 mole of Li2CO3. So, the mass of Li2CO3 formed will be the same as the mole ratio:

0.903 mol Li2CO3 x 40.1 g/mol Li2CO3 = 36.2 g of Li2CO3

Therefore, by adding 37g of LiOH to the reaction, approximately 36.2g of Li2CO3 will be produced.

1. To balance the equation:

The first step is to count the number of atoms for each element on both sides of the equation. In this case, we have:

Left side:
1 Li, 1 O, 1 H, 1 C

Right side:
2 Li, 3 O, 1 C, 1 H

The next step is to adjust the coefficients in front of each compound to equalize the number of atoms on both sides.

The balanced equation is:

2 LiOH + CO2 ---> Li2CO3 + H2O

2. The stoichiometric ratios for Li2CO3 can be determined by inspecting the coefficients in the balanced equation.

From the balanced equation:
2 moles of LiOH will react with 1 mole of CO2 to produce 1 mole of Li2CO3.

This means that for every 2 moles of LiOH you have, you will produce 1 mole of Li2CO3.

3. To find the mass of Li2CO3 formed when adding 37g of LiOH to the reaction, you need to use stoichiometry and the molar mass of Li2CO3.

Step 1: Calculate the number of moles of LiOH:
moles = mass / molar mass
moles of LiOH = 37g / (6.94 g/mol + 15.999 g/mol + 1.008 g/mol)
moles of LiOH ≈ 0.700 moles

Step 2: Use the stoichiometric ratio to calculate the moles of Li2CO3 produced:
moles of Li2CO3 = 0.700 moles of LiOH * (1 mole of Li2CO3 / 2 moles of LiOH)
moles of Li2CO3 ≈ 0.350 moles

Step 3: Calculate the mass of Li2CO3:
mass = moles * molar mass
mass of Li2CO3 = 0.350 moles * (6.94 g/mol + 12.011 g/mol + (3 x 15.999 g/mol))
mass of Li2CO3 ≈ 10.549g (rounded to three decimal places)

Therefore, approximately 10.549 grams of Li2CO3 will be formed when 37 grams of LiOH react in the given reaction.