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!
The mass for of aluminum that is produced by the decomposition of 5.0 Kg Al2O3 is 2647 g or 2.647 Kg
calculation
Write the equation for decomposition of Al2O3
Al2O3 = 2Al + 3 O2
find the moles of Al2O3 = mass/molar mass
convert 5 Kg to g = 5 x1000 = 5000 grams
molar mass of Al2O3 = 27 x2 + 16 x3 = 102 g/mol
moles =5000 g/ 102 g/mol = 49.0196 moles
by use of mole ratio between Al2O3 to Al which is 1:2 the moles of Al = 49.0196 x2 =98.0392 moles
mass of Al = moles x molar mass
= 98.0392 moles x 27g/mol = 2647 grams or 2647/1000 = 2.647 Kg
Run it up by nav and lemonade by dont oliver
