Question

A student carefully placed 23.0 g of sodium in a reactor with an excess quantity of chlorine gas. When the reaction is complete, the student obtained 58 grams of salt. How many grams of sodium reacted?

Answer 1

Approximately 23.0 grams of sodium reacted with chlorine gas to produce 58 grams of sodium chloride, as calculated using the stoichiometry of the balanced chemical equation and the molar masses of sodium and sodium chloride.

Step-by-step explanation:

To determine how many grams of sodium reacted with chlorine gas to produce sodium chloride, we need to understand reaction stoichiometry. According to the balanced chemical equation 2 Na(s) + Cl₂(g) → 2 NaCl(s), 2 moles of sodium react with 1 mole of chlorine gas to produce 2 moles of sodium chloride. We can calculate the mass of sodium that reacted by using the mass of sodium chloride produced and the molar masses of sodium and sodium chloride.

First, calculate the moles of sodium chloride produced using its formula mass (58.44 amu):

Moₓ(NaCl) = 58 g / 58.44 g/mol ≈ 1 mol

According to the stoichiometry, 1 mole of sodium chloride comes from 1 mole of sodium; therefore, the mass of sodium that reacted is:

Mass of Na = Moles of Na × Molar mass of Na ≈ 1 mol × 22.99 g/mol = 22.99 g

Thus, approximately 23.0 grams of sodium reacted (which matches the initial amount placed in the reactor, indicating all sodium reacted).

Answer 2

Answer: 22.8 grams

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} NaCl=(58g)/(58.5g/mol)=0.99moles`

The balanced chemical reaction is :

`2Na+Cl_2\rightarrow 2NaCl`

According to stoichiometry :

As 2 moles of
`NaCl` is produced by = 2 moles of
`Na`

Thus 0.99 moles of
`NaCl` is produced by =
`(2)/(2)* 0.99=0.99moles` of
`Na`

Mass of
`Na=moles* {\text {Molar mass}}=0.99moles* 23g/mol=22.8g`

Thus 22.8 g of
`Na` is reacted.