<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
Increase in heat, molecules start to escape and it turns to vapor
1 milliliters in one milligram.
To determine molecular formula, we first need to find out its empirical formula,
Carbon. Hydrogen. Nitrogen. Oxygen
Mass. 49.98g. 5.19g. 28.85g. 16.48g
Mole. 4.165. 5.19. 2.06. 1.03
Divide 4. 5. 2. 1
by
smallest
So by comparing the mole ratio from the table above, i hope u understand the table
The empirical formula is C4H5N2O
given molecular mass = 194.19g
so
(C4H5N2O) n= 194.19
(48+5+28+16)n=194.19
n= 2
molecular formula = C8H10N4O2
