Sorry, I was slightly confused. Now I have found there to be .266 g of Br and 1.05 g of Cl, matching the total mass. Thank you for the help so far, but I have one final question. How do I determine the mass of NaBr from the mass of Na and Br when Na is split into two different molecules?
I did: \frac{.76 g Na}{1} x \frac{mol Na}{22.99 g} = .033 mol Na
2.08 gmixture - .76 gNa = 1.32 gCl + Br
I then tried: \frac{.033 mol Na}{1} x \frac{mol Br}{2 mol Na} x \frac{79.909 g}{mol Br} = 1.32 g Br
That mass of Br cannot be true, so apparently I am doing something wrong.
Homework Statement
A mixture of NaCl and NaBr has a mass of 2.08 g and is found to contain 0.76 g of Na. What is the mass of NaBr in the mixture?
Homework Equations
Molar mass equations (possibly), formula mass
The Attempt at a Solution
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Homework Statement
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Homework Statement
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Homework Equations
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