For the reaction 2Na(s) + Cl(g) �¨2NaCl(s), how many grams of NaCl could be produced from 103.0 g of Na and 13.0 L of Cl (at STP)?

How many moles of Cl2 is 13L at stp?

How many moles of Na is 103g?

It appears that Cl2 is the limiting reactant, so for each mole of Cl2, you get two moles of NaCl. Take the moles of the Cl2 (if it is indeed the limiting reactant) and multiply it by 2, then convert that to grams of NaCl

To solve this problem, we need to use stoichiometry and the concept of molar ratios.

Step 1: Convert grams of Na to moles of Na.
We know that the molar mass of Na is 22.99 g/mol.
Therefore, moles of Na = grams of Na / molar mass of Na
= 103.0 g / 22.99 g/mol
≈ 4.48 mol Na

Step 2: Convert volume of Cl to moles of Cl.
We need to use the ideal gas law to convert the volume of Cl gas to moles. At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 L.
Therefore, moles of Cl = volume of Cl / 22.4 L/mol
= 13.0 L / 22.4 L/mol
≈ 0.580 mol Cl

Step 3: Determine the limiting reactant.
In this reaction, two moles of Na react with one mole of Cl to produce two moles of NaCl.
The molar ratio between Na and Cl is 2:1.
Since we have fewer moles of Cl than the stoichiometric ratio requires, Cl is the limiting reactant.

Step 4: Use stoichiometry to find the moles of NaCl produced.
From the balanced chemical equation, we know that 2 moles of Na react to form 2 moles of NaCl. Therefore, 1 mole of Na reacts to form 1 mole of NaCl.
Hence, moles of NaCl produced = moles of Na
= 4.48 mol Na

Step 5: Convert moles of NaCl to grams of NaCl.
The molar mass of NaCl is 58.44 g/mol.
Therefore, grams of NaCl produced = moles of NaCl * molar mass of NaCl
= 4.48 mol NaCl * 58.44 g/mol
≈ 261.63 g NaCl

So, approximately 261.63 grams of NaCl could be produced from 103.0 grams of Na and 13.0 liters of Cl at STP.

To calculate the number of grams of NaCl that can be produced, we need to determine the limiting reactant and then use stoichiometry to find the amount of NaCl formed.

Step 1: Convert the given quantities to moles.
- The molar mass of Na is 22.99 g/mol.
- The molar volume of any gas at STP is 22.4 L/mol.
- The molar mass of Cl is 35.45 g/mol.

103.0 g of Na can be converted to moles using the molar mass:
moles of Na = 103.0 g / 22.99 g/mol

13.0 L of Cl can be converted to moles using the molar volume at STP:
moles of Cl = 13.0 L / 22.4 L/mol

Step 2: Determine the limiting reactant.
The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product formed. To find the limiting reactant, we need to compare the moles of each reactant with the stoichiometric ratio in the balanced equation.

From the balanced equation, 2 moles of Na react with 1 mole of Cl to produce 2 moles of NaCl.

Calculate the moles of NaCl that can be produced from the moles of Na and Cl:
moles of NaCl from Na = (moles of Na) / 2
moles of NaCl from Cl = (moles of Cl) / 1

The limiting reactant is the reactant that produces fewer moles of NaCl. Compare the two quantities obtained above to find the limiting reactant.

Step 3: Calculate the moles of NaCl produced.
The moles of NaCl produced will be equal to the moles of NaCl from the limiting reactant.

Step 4: Convert moles of NaCl to grams of NaCl.
The molar mass of NaCl is 58.44 g/mol.
grams of NaCl = (moles of NaCl) × (molar mass of NaCl)

By following these steps, you can determine the number of grams of NaCl that can be produced from 103.0 g of Na and 13.0 L of Cl at STP.