Question

Tion 1 of 20

Answer 1

Answer:

None of these

Explanation:

The questions asks us to find the amount of sodium chloride formed (in grams) when 1.125 X 10^45 molecules of chlorine gas reacts with sodium.

We start by converting the molecules of chlorine gas to moles using Avogadro's number:

1.125 X 10^45 molecules Cl2 X 1 mol Cl2 / 6.022 X 10^23 = 1.87 X 10^21 mol

From here we can set up a BCA table, assuming that we have excess Sodium to react with: We have to assume that Cl2 is the limiting reactant.

       2Na          +           Cl2          ---->         2NaCl

B .  excess             1.87 X 10^21                       0

C -3.74 X 10^21     -1.87 X 10^21              + 3.74 X 10^21

A   X Amount                  0                          3.74 X 10^21

Now we have 3.74 X 10^21 moles of NaCl that we have to convert to grams:

We can use the molar mass of sodium chloride to accomplish this:

3.74 X 10^21 moles NaCl X 58.44 g NaCl / 1 mol NaCl = 2.18 X 10^23 g NaCl

Answer: None of these

Answer 2

Answer:

The mass of NaCl produced is 1.2 * 10^23 grams

Explanation:

Step 1: Data given

Number of molecules of Cl2 - gas = 1.25 * 10^45 molecules

Number of Avogadro = 6.022 * 10^23

Step 2: The balanced equation

2 Na + Cl2 → 2 NaCl

Step 3: Calculate moles of Cl2

Moles Cl2 = 1.25 *10^45 / 6.022*10^23

Moles Cl2 = 2.1 *10^21 moles

Step 4: Calculate moles NaCl

For 2.1 *10^21 moles Na we'll have 2.1*10^21 moles NaCl

Step 5: Calculate mass NaCl

Mass NaCl =  moles NaCl * molar mass NaCl

Mass NaCl = 2.1*10^21 moles * 58.44 g/mol

Mass NaCl = 1.2 *10^23 grams

The mass of NaCl produced is 1.2 * 10^23 grams