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If we react 5.4g of sodium chloride with an unknown amount of fluorine gas, we produce 4.9g of sodium fluoride and 3.7g chlorine gas. How much fluorine was consumed in the reaction?

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Answer:

1.9g of F₂ were consumed in the reaction

Step-by-step explanation:

Based on the chemical reaction:

2NaCl + F₂ → 2 NaF + Cl₂

Where 2 moles of NaCl react per mole of F₂ to produce 2 moles of NaF and 1 mole of Cl₂

To understand in a better way how the reaction occurs we need to convert each mass of compound to moles:

Moles of NaCl -Molar mass: 58.44g/mol-:

5.4g * (1mol / 58.44g) = 0.092 moles NaCl

Moles of NaF -Molar mass: 41.988g/mol-:

4.9g * (1mol / 41.998g) = 0.12 moles NaF

Moles of Cl₂ -Molar mass: 70.90g/mol-:

3.7g * (1mol / 70.90g) = 0.052 moles Cl₂

That means approximately 0.1 moles of NaCl react with 0.05 moles of F₂ to produce 0.1moles of NaF and 0.05 moles of Cl₂.

0.05 moles of F₂ are:

0.05 moles * (38g/mol) =

1.9g of F₂ were consumed in the reaction

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