For the reaction
2Cl2(g)+4NaOH(aq)->3NaCl(aq)+ NaClO2(aq)+2H2O(l)
how many grams of NaCl can be produced from 30.8 g of Cl2 and excess NaOH?
To find the grams of NaCl produced from 30.8 g of Cl2, we first need to determine the limiting reactant. The limiting reactant is the one that is completely consumed, and it determines the maximum amount of product that can be formed.
Step 1: Convert grams of Cl2 to moles.
The molar mass of Cl2 is 35.45 g/mol.
30.8 g Cl2 * (1 mol Cl2 / 35.45 g Cl2) = 0.868 mol Cl2
Step 2: Use the stoichiometric ratio to calculate the moles of NaCl formed.
From the balanced equation, we see that the stoichiometric ratio between Cl2 and NaCl is 2:3.
0.868 mol Cl2 * (3 mol NaCl / 2 mol Cl2) = 1.30 mol NaCl
Step 3: Convert moles of NaCl to grams.
The molar mass of NaCl is 58.44 g/mol.
1.30 mol NaCl * (58.44 g NaCl / 1 mol NaCl) = 75.77 g NaCl
Therefore, you can produce 75.77 grams of NaCl from 30.8 grams of Cl2 and excess NaOH.
To find the amount of NaCl that can be produced from 30.8 g of Cl2, we need to perform a stoichiometric calculation.
Step 1: Find the molar mass of Cl2
The molar mass of Cl2 is 35.45 g/mol x 2 = 70.90 g/mol.
Step 2: Convert grams of Cl2 to moles
Given that we have 30.8 g of Cl2, we divide it by the molar mass of Cl2:
30.8 g / 70.90 g/mol = 0.434 mol
Step 3: Apply the stoichiometry
From the balanced equation, we see that the molar ratio between Cl2 and NaCl is 2:3. That means for every 2 moles of Cl2, we get 3 moles of NaCl.
Since we have 0.434 moles of Cl2, we can calculate the moles of NaCl:
0.434 mol Cl2 x (3 mol NaCl / 2 mol Cl2) = 0.651 mol NaCl
Step 4: Convert moles of NaCl to grams
Finally, we convert moles of NaCl to grams by multiplying by the molar mass of NaCl, which is 58.44 g/mol:
0.651 mol NaCl x 58.44 g/mol = 38.0 g NaCl
Therefore, from 30.8 g of Cl2, we can produce approximately 38.0 grams of NaCl.
Worked example. Follow the steps.
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