A sample of sodium reacts completely with 142 g of chlorine, forming 234 g of sodium chloride. What mass of sodium reacted?
To determine the mass of sodium that reacted, we need to use the given information and apply the concept of stoichiometry.
First, let's write down the balanced equation for the reaction between sodium and chlorine:
2Na + Cl2 -> 2NaCl
From the equation, we can see that 2 moles of sodium react with 1 mole of chlorine to produce 2 moles of sodium chloride.
Next, let's find the number of moles of sodium chloride formed. We know that the molar mass of sodium chloride (NaCl) is 58.44 g/mole (22.99 g/mol for sodium + 35.45 g/mol for chlorine).
Given that the mass of sodium chloride formed is 234 g, we can calculate the number of moles using the formula:
moles = mass / molar mass
moles = 234 g / 58.44 g/mol
moles ≈ 4 moles
Since the mole ratio between sodium and sodium chloride is 2:2, we can conclude that 4 moles of sodium reacted.
Finally, let's determine the mass of sodium that reacted using the mole-to-mass conversion:
mass = moles × molar mass
mass = 4 moles × 22.99 g/mol (molar mass of sodium)
mass ≈ 91.96 g
Therefore, approximately 91.96 grams of sodium reacted.