CHEM reacting mass 計算問題,急！！！

23.
b.
4g Mg = 4/24.3 = 0.164 mol Mg
3.2g O2 = 3.2/32 = 0.1 mol O2
since mole ratio of the equation is Mg : O2 = 2 : 1
reacting mol: Mg<O2
Mg is the limiting agent

c.
mol of O2 reacted = 0.164/2 = 0.082 mol O2
O2 not reacted = 0.1 - 0.082 =0.018 mol O2 =0.018* 32 = 0.576g

d.
mass of MgO given out
=4+3.2-0.576=6.624g

24.d.
3g CuO = 3/(76.55)=0.038 mol CuO
3g C = 3/12=0.25 mol C
since mole ratio of the equation is CuO : C = 2 : 1
so CuO is the limiting agent

e.
CuO reacted = 0.038 mol
so 0.038 mol Cu is formed
0.038 mol Cu = 2.41g Cu

25.a.
CaCO3(s) + 2 HCl(aq) => CO2(g) + H2O(l) + CaCl2(q)

b.i.
0.15g of CO2 gas is given out
0.15g CO2 = 0.15/(44)=0.0034 mol CO2

ii.
reacting ratio of CO2 : CaCO3 = 1:1
so 0.0034 mol CaCO3 is in the chalk
mass of CaCO3 in the chalk = 0.0034*100=0.34
%by mass of CaCO3 = 0.34/0.4*100% = 85%

c.The chalk may not react competely

26.b
45.69-40.89=4.8g

c.
mol of the Mg ribbon = 40.89-37.27 = 36.2g Mg = 1.489 mol Mg
mol of O reacted = 4.8/1.6=3 mol
mol of Mg : mol of O = 1.5:3 = 1:2

in calculation of "limiting reagent" / "excess reagent", first write the equations (you did it). then, calculate the no. of moles of EACH reagent. next, according to the equation and their mole ratio, see which one is in excess. i'll show you by Q.23.

Q. 23
no. of mole of Mg = mass / molar mass
= 4.00/24.3 = 0.1646 mole
no. of mole of oxygen = mass / molar mass
= 3.2/(16x2) = 0.1 mole

mole ratio of Mg:O2 = 2:1
for Mg to complete reaction, no. of mole of oxygen needed = 0.1646/2 = 0.0823 mole (< 0.1 mole).
Thus magnesium is limiting agent; oxygen is in excess by: 0.1 - 0.0823 = 0.0177 mole
mass of oxygen in excess = 0.0177 x 32 = 0.5664 g

mole ratio of Mg:MgO = 1:1
thus no. of mole of MgO = 0.1646 mole
mass = no. of mole x molar mass = 0.1646 x (24.3+16) = 6.633 g

SEE how to deal with this kind of Q.?

For Q. 25-b-i,
you have mass of CO2, and by no. of mole = mass / molar mass,
you should be able to calc. no. of mole of CO2
b-ii:
mole ratio of CaCO3:CO2 = 1:1
after calculating no of mole of CaCO3, turn it into mass: = no. of mole x molar mass
Purity = (actual mass of analyte in sample) / (total mass of sample)

error? some compound else can react with HCl and give CO2, but is not CaCO3... eg MgCO3.

Q.26
write the equation of reaction first!!
after ignition, increase in mass comes from air -- oxygen. calculate mass of oxygen, thus its no. of mole

mass of magnesium is given (... - mass of empty crucible), calculate no. of mole of Mg
with two no. of mole, you can get the mole ratio of Mg:O.


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