For a and b, you need to divide it by Avogadro’s number to find the answer.
a. (6.022x10^23)/6.022x10^23 = 1 mole of Ne
b. (3.011x10^23)/6.022x10^23 = 0.5 moles of Mg
For c and d, you’ll use the mass provided divided by the molar mass to find the number of moles.
Pb molar mass = 207.2 g/mol
c. (3.25x10^5)/207.2 = 1.57x10^3 moles of Pb
For d, I can’t tell if is Cu, C or something else but you can follow the steps above to solve the problem.
Answer: a. 7.31 g
b. 20.4 %
Explanation:
To calculate the moles :
According to stoichiometry :
3 moles of require = 1 mole of
Thus 0.645 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 3 moles of give = 2 moles of
Thus 0.645 moles of give = of
Mass of
Thus theoretical yield for this reaction under the given conditions is 7.31 g.
b)
The percent yield for this reaction under the given conditions is 20.4 %
Answer:
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I think you forgot to post a picture