Homework help!!?
A student in lab performs Reaction B with 1.468 g of sodium carbonate reactant, and ends up collecting 1.340 g of sodium chloride product. What is her percent yield?
the equation is experimental value/theoretical value x 100
but what is my experimental and theoretical value??!!
adshfjlt this is so confusing.
thanx for the help!!
Answer:
We don't have the complete reaction here, but we might be able to piece things together.
Sodium carbonate is Na2CO3. The "2" is key.
Figure out how many moles of that you start with (1.146 g divided by 106 g/mol = 0.0138 mol of Na2CO3).
Sodium chloride is NaCl. There's only ONE sodium in NaCl, so you should get TWO moles of sodium chloride for every mole of Na2CO3 you started with. 0.0138 x 2 = 0.0277 moles of NaCl expected, which corresponds to 0.0277 moles x 58 g/mol = 1.61 g of NaCl expected.
So, THEORETICALLY, you should get 1.61 g. But EXPERIMENTALLY you got 1.340.
1.340 / 1.61 x 100 = 83%.
One simple way to check your work on this if you get confused is that your percent yield is never higher than 100.
ex./theoretical x 100
first, you need to find your experimental value.
You need an equation though. sodiumCarbonate is Na2CO3 because CO3 is a -2charge and Na is a +1 charge, you need two Na's.
Na2CO3 ------> NaCl
1.46g 1.34g
Use limiting reagent method to determine which one is the limiting amount. This will be your accepted value. Your experimental are the values they gave. Hope this helps :)
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