Question

1, 2 and 3 Onyx D 1, 3 and 4 only x (B) 14. Sodium reacts with chlorine gas to form sodium chloride according to the equation below.

2Na(s) + Cl2(g) → 2NaCl(s)

If 0.02 mol of sodium was reacted with 300 cm? of chlorine gas, what is the volume of excess chlorine gas that remained at the end of the reaction?
А 60 cm3
B 30 cm3
C 120 cm3
D 240 cm3 ​

Answer 1

[tex]Explanation:

Start with a balanced equation.

2Na(s) + Cl2(g)→2NaCl(s)

In order to determine the mass of chlorine gas that reacted to produce 234 g NaCl, you work backwords, starting with NaCl and ending with Na.

The process will involve the following steps:

given mass NaCl→mol NaCl→mol Na→mass Na

Determine the mol NaCl that reacted.

Multiply the given mass of NaCl by the inverse of its molar mass.

The molar mass of NaCl is 58.44 g/mol NaCl.

234g NaCl×1mol NaCl58.44g Cl2=4.004 mol NaCl

Determine the moles of Na that reacted.

Multiply mol NaCl by multiplying mol NaCl by the mol ratio between NaCl and Na from the balanced equation, so that mol NaCl is cancelled.

4.004mol NaCl×2mol Na2mol NaCl=4.004 mol Na

Determine the mass of Na that reacted.

Multiply mol Na by its molar mass.

The molar mass of Na is 22.98976928 g/mol Na.(Periodic table)

4.004mol Na×22.98976928g Na1mol Na=92.1 g Na=0.0921 kg(rounded to three significant numbers)

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