Answer:
Honestly, I'm not even sure what the incorrect student was trying to do. The correct answer is:
( ) = the number of elements
(1)N2 and (3)H2
I think the student added the N2 and H2 to "balance" the chemical equation when instead they should have tried the balance the equation by making sure the atoms on the reactant side are equal to the atoms on the product side. You do this by multiplying the atoms on the reactant side until they equal the product side (not by adding them, as the student did).
It got planted and went into the process of the specific foot item
To answer your question I will use dimensional analysis, which is used by cancelling out the units. I will also use the balanced equation provided as a conversion factor.
A) First start out with the 0.300 mol of C6H12O6...
0.300 mol C6H12O6 * (2 mol CO2 / 1 mol C6H12O6) = 0.600 mol CO2
*The significant figures (sig figs) at still three, the 2 is a conversion counting number and does not count*
B) First change 2.00 g of C2H5OH to moles of C2H5OH...
The molecular mass of C2H5OH is...
2(12.01 g/mol) + 5(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol = 46.07 g/mol
This can be used as a conversion factor to change grams to moles.
2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) = 0.0434 mol C2H5OH
Second, you can change the moles of C2H5OH to moles of C6H12O6..
0.0434 mol C2H5OH * (1 mol C6H12O6 / 2 mol C6H12O6) = 0.0217 mol C6H12O6
Third, change moles of C6H12O6 to grams...
MM = 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol) = 180.16 g/mol
0.0217 mol C6H12O6 * (180.16 g C6H12O6 / 1 mol C6H12O6) = 3.91 g C6H12O6
C) Now I am going to put it all into one long dimensional analysis problem.
MM of CO2 = 44.01 g/mol
MM of C2H5OH = 46.07 g/mol
2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) * (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2
I hope this helped and I am sorry that I talked to much, I just didn't want to miss anything!
Answer:
I don't know how can i do
Explanation:
please give me hint
