A. M x L = moles.
<span>b. CH3COOH + NaOH ==> CH3COONa + H2O </span>
<span>I...6 mmols....0.......7.5 mmoles </span>
<span>C... 0........0.51 mmols..0 </span>
<span>E...6-0.511 ....0.......7.5+0.511 </span>
<span>I stands for initial </span>
<span>C stands for change. </span>
<span>E stands for equilibrium. </span>
<span>Just divide mmoles by 1000 to convert to moles. I work in mmoles because I get tired of writing those zeros. </span>
<span>c. done as in b.</span>
<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>
