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# How many liters of fluorine gas can react with 32.0 grams of sodium metal at standard temperature and pressure?

How many liters of fluorine gas can react with 32.0 grams of sodium metal at standard temperature and pressure? Show all of the work used to find your answer.

2Na + F2 yields 2NaF

A laboratory experiment requires 2.0 L of a 1.5 M solution of hydrochloric acid (HCl), but the only available HCl is a 12.0 M stock solution. How could you prepare the solution needed for the lab experiment? Show all the work used to find your answer.

### 2 Answers

- pisgahchemistLv 79 years agoFavorite Answer
2Na(s) + F2(g) --> 2NaF(s)

32g .......??? L (STP)

1. Convert the mass of Na to moles of Na

2. Use the mole ratio from the balanced equation to find moles of F2

3. Convert the moles of F2 to liters of F2 at STP

32 g Na x (1 mol Na / 23.0 g Na) x (1 mol F2 / 2 mol Na) x (22.4 L / 1 mol F2) = 15.6 L of fluorine

For dilutions use the following equation:

Md x mLd = Mc x mLc ..... where the subscript "d" refers to the dilute solution and "c", the concentrated. Solve for mLc to get the volume of concentrated HCl that must be diluted. Place it in a 2L graduated cylinder and dilute to the 2L mark. (Well, usually it's milliliters... or it can be liters:

Md x Ld = Mc x Lc

============ Follow up ============

The atomic weight of sodium is 22.9898 amu. Therefore, rounding it to 1 decimal point will make it 23.0, not 22.9 as Hellamund has suggested. I've taught AP chemistry for nearly 40 years, and I tell my students that they are inviting round-off errors by rounding at each step. This is why I prefer to do all the steps in sequence (called a "chain calculation") as shown above. Do the calculations in the calculator as a chain, and then you round to the correct number of significant digits at the end.

- ChopkinsCafeLv 49 years ago
1)

Convert grams of sodium to moles of sodium using its molar mass of 22.9 g/mol.

32.0 g * (1 mol/22.9 g) = 1.40 mol

Calculate how many moles of fluorine are required to react with 1.40 mol of sodium.

1.40 mol Na * (1 F2/2 Na) = 0.700 mol F2

At STP (standard temperature and pressure), 1 mol gas = 22.4 L gas

0.700 mol * (22.4 L/1 mol) = 15.7 L

2)

Use the dilution formula, M1V1=M2V2, where M is molarity and V is volume.

1.5*2.0 = 12.0*V

V = 0.25 L

This means that you need 0.25 L of the 12.0 M HCl to have the same number of moles of HCl as in the 2.0 L of 1.5 M HCl. But of course you need to add the 0.25 L of 12.0 M HCl into 1.75 L of water so that it is 2.0 L.

Source(s): AP chem