A sample of sodium reacts completely with 0.213 kgkg of chlorine, forming 351 gg of sodium chloride. What mass of sodium reacted?

0.138 kg of Na.

Explanation:

Equation of the reaction

2Na (aq) + Cl2 (g) - -> 2NaCl (s)

By stoichiometry, since 2 moles of Na reacted with 1 mole of Cl2 to produce 2 moles of NaCl. Therefore,

Calculating the number of moles present,

Cl2:

Number of moles = mass/molar mass

Molar mass of Cl2 = 35.5 * 2

= 71 g/mol.

= 213 * 71

= 3 moles.

NaCl:

Molar mass = 23 + 35.5

= 58.5 g/mol.

Number of moles = mass/molar mass

= 351/58.5

= 6 moles.

Since they both have equal moles,

Mass of Na = number of moles * molar mass

Molar mass of Na = 23 g/mol

= 23 * 6 moles

= 138 g

= 0.138 kg of Na.

The mass of sodium reacted is 138 grams

Explanation:

Firstly, the chemical reaction should be represented with a balance chemical equation. The chemical equation can be written as follows.

Na = sodium

Cl2 = chlorine gas

Na + Cl2 → NaCl

The balanced equation is

2Na (s) + Cl2 (g) → 2NaCl (aq)

Atomic mass of sodium = 23 grams per mol

Atomic mass of chlorine = 35.5 grams per mol

From the chemical equation the molar mass are as follows

Sodium = 2 * 23 = 46 grams

Chlorine = 35.5 * 2 = 71 grams

Sodium chloride = 2 * 23 + 2 * 35.5 = 117 grams

if 46 grams sodium reacted to produce 117 grams of sodium chloride

? grams of sodium react to produce 351 grams sodium chloride

grams of sodium reacted = (351 * 46) / 117

grams of sodium reacted = 16146/117

grams of sodium reacted = 138 grams