Solutions of sodium hypochlorite, NaClO are sold as bleach. They are prepared by the following reaction:
2NaOH + Cl2 --> NaCl + NaClO + H2O
If you have 2.32 mol of NaOH and 2.74 mol of Cl2 ga available to react, answer the following questions
Na: 23.0 amu Cl:35.5 amu O: 16.0 amu
1 How many moles of NaClO can be produced from the sodium hydroxide?
2 How many moles of NaClO can be produced from the chlorine gas?
3 Based on your findings for a and b, what is the maximum number of grams of sodium hypochlorite that can be produced from the moles of sodium hydroxide and chlorine gas provided?
To solve this problem, we need to use the stoichiometry of the reaction.
1) To determine the number of moles of NaClO produced from the sodium hydroxide (NaOH), we need to find the stoichiometric ratio between NaOH and NaClO.
From the balanced equation, we can see that for every 2 moles of NaOH, 1 mole of NaClO is produced.
Therefore, the number of moles of NaClO produced from 2.32 mol of NaOH can be calculated as follows:
Number of moles of NaClO = (2.32 mol NaOH) / (2 mol NaOH: 1 mol NaClO)
= 1.16 mol NaClO
So, 1.16 moles of NaClO can be produced from 2.32 mol of NaOH.
2) To determine the number of moles of NaClO produced from the chlorine gas (Cl2), we need to find the stoichiometric ratio between Cl2 and NaClO.
From the balanced equation, we can see that for every 1 mole of Cl2, 1 mole of NaClO is produced.
Therefore, the number of moles of NaClO produced from 2.74 mol of Cl2 can be calculated as follows:
Number of moles of NaClO = (2.74 mol Cl2) / (1 mol Cl2: 1 mol NaClO)
= 2.74 mol NaClO
So, 2.74 moles of NaClO can be produced from 2.74 mol of Cl2.
3) To determine the maximum number of grams of sodium hypochlorite (NaClO) that can be produced from the moles of sodium hydroxide (NaOH) and chlorine gas (Cl2) provided, we need to use the molar mass of NaClO.
The molar mass of NaClO = 23.0 g/mol (Na) + 35.5 g/mol (Cl) + 16.0 g/mol (O) = 74.5 g/mol.
To calculate the maximum number of grams, we can use the formula:
Mass of NaClO = Number of moles of NaClO × Molar mass of NaClO.
For the moles of NaClO produced from NaOH, the mass can be calculated as:
Mass of NaClO = 1.16 mol NaClO × 74.5 g/mol NaClO = 86.12 g NaClO (rounded to two decimal places).
For the moles of NaClO produced from Cl2, the mass can be calculated as:
Mass of NaClO = 2.74 mol NaClO × 74.5 g/mol NaClO = 204.05 g NaClO (rounded to two decimal places).
Therefore, the maximum number of grams of sodium hypochlorite that can be produced from the moles of sodium hydroxide and chlorine gas provided is 86.12 g NaClO from NaOH and 204.05 g NaClO from Cl2.
To solve these questions, we will need to use stoichiometry, which involves the balanced chemical equation and mole ratios.
Let's start by calculating the number of moles of NaClO that can be produced from the sodium hydroxide (NaOH).
1) First, we need to determine the limiting reactant. This is the reactant that will be completely consumed and will limit the amount of product formed.
To find the limiting reactant:
- Convert the given moles of NaOH to moles of NaClO using the mole ratio from the balanced equation:
2 moles of NaOH : 1 mole of NaClO
Given: 2.32 mol NaOH
Mole ratio: 2 moles NaOH : 1 mole NaClO
So, (2.32 mol NaOH) x (1 mole NaClO / 2 moles NaOH) = 1.16 mol NaClO
Therefore, 1.16 moles of NaClO can be produced from 2.32 moles of NaOH.
2) Now, let's determine the number of moles of NaClO that can be produced from the chlorine gas (Cl2).
To find the limiting reactant:
- Convert the given moles of Cl2 to moles of NaClO using the mole ratio from the balanced equation:
1 mole of Cl2 : 1 mole of NaClO
Given: 2.74 mol Cl2
Mole ratio: 1 mole Cl2 : 1 mole NaClO
So, 2.74 mol Cl2 can produce 2.74 mol NaClO.
Therefore, 2.74 moles of NaClO can be produced from 2.74 moles of Cl2.
3) Based on the findings for parts 1 and 2, we need to determine the maximum number of grams of sodium hypochlorite (NaClO) that can be produced from the moles of sodium hydroxide and chlorine gas provided.
To calculate the maximum amount of NaClO:
- Multiply the moles of NaClO formed from NaOH by the molar mass of NaClO.
Given: Moles of NaClO from NaOH = 1.16 mol
Molar mass of NaClO = (23.0 amu + 35.5 amu + 16.0 amu) g/mol = 74.5 g/mol
So, (1.16 mol NaClO) x (74.5 g NaClO / 1 mol NaClO) = 86.32 g NaClO
Thus, the maximum number of grams of sodium hypochlorite that can be produced from 2.32 moles of NaOH is 86.32 grams.
Since the moles of NaClO from chlorine gas (2.74 mol) are greater than the moles of NaClO from NaOH (1.16 mol), the moles of NaClO from chlorine gas (2.74 mol) will determine the maximum amount of NaClO that can be produced.
Therefore, the maximum number of grams of sodium hypochlorite that can be produced from the given amount of NaOH and Cl2 is 86.32 grams.