Given the unbalanced equation a. Na + CI_2 rightarrow NaCl Balanced the equation
ID: 961386 • Letter: G
Question
Given the unbalanced equation a. Na + CI_2 rightarrow NaCl Balanced the equation. If 8.3 g of Na and 14.0 g of Cl_2 are reacted together in a lab experiment: b. Calculate the number of moles of each reactant, based on lab data c. What is the limiting reagent? d. Calculate the number of moles of sodium required to react with all the chlorine present, based on the balanced equation. e. What is the maximum number of moles of NaCl that can be formed? f. What is the maximum number of grams of NaCl that can be formed? 2. If 16.8 grams of copper is placed into a solution containing 25.65 g of silver nitrate, what mass of silver will be produced? (The other product is copper (II) nitrate.) Write the balanced equation. 3. If 19.8 g of ammonium phosphate is mixed in solution with 12.4 g of calcium nitrate, what mass of precipitate is produced? Write the balanced equation. 4. For the reaction PCl_3 + Cl_2, 16.0 g of phosphorus trichloride is reacted with 13.2 g of chlorine gas. What is the maximum mass of phosphorus pentachloride that could be produced?Explanation / Answer
1. a) The equation is well balanced, let's write again:
2Na + Cl2 ------> 2NaCl
b) moles Na = 8.3 / 23 = 0.3609 moles
moles Cl = 14 / 70.9 = 0.1975 moles
c) for the limiting reagent:
2 moles of Na ----> 1 mol Cl2
0.3609 ---------> x
x = 0.3609 / 2 = 0.1805 moles of Cl2
And we have 0.1975 moles of Cl2, this means that the Cl2 is in excess and the Na is the limiting reactant.
d) for the moles of sodium, we do the same thing as part c but with cl:
0.1975 * 2 = 0.395 moles of Na required to react completely with the moles of Cl
e) with the limiting reactant of Na, the maximun moles of NaCl, by stechiometry, moles of Na = moles NaCl so:
moles NaCl = 0.3609 moles
And assuming the result of part d) the maximum number of moles that can be formed it would be 0.395 moles NaCl.
f) for the mass:
mNaCl = 0.3609 * (23+35.45) = 21.09 g
And assuming result of part d)
mNaCl = 0.395 * (23+35.45) = 23.09 g.
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