1) If 2.00 g of oxygen gas contains xmolecules, how many molecules are present in 28.0 g of nitrogen gas? (Relative atomic masses: N = 14.0, O = 16.0)2) Asolid mixture of iron(II) nitrate and iron(II) carbonate contains 1.0 mole ofnitrate ions and 1.2 moles of iron(II)ions. What is the number of moles ofcarbonate ions in the mixture?
3)Considerthe following equation:X2(g) + 3Y2(g) à 2XY3(g)
If 4 moles of X2(g) react with 6 moles of Y2(g), what is the number of moles of XY3(g) formed?
4) Ifthere are x molecules in 17.75 g of chlorine, how many molecules are present in8.00 g of sulphur dioxide? (Relativeatomic masses: O = 16.0, S = 32.1, Cl = 35.5) 5)A sample of paint pigment of mass 1.50 g is dissolved and the lead ions in it are separated by precipitation as solid PbSO4. The precipitate has a mass of 0.0849 g. What is the percentage by mass of lead in the pigment? (Relative atomic masses: O = 16.0, S = 32.1, Pb = 207.2)
F4 chem mole calculation
2011-01-16 10:55 pm
回答 (2)
2011-01-17 2:20 am
✔ 最佳答案
obviously this's your homework. you should show your understanding and interpretation on the questions before asking (directly for answers). don't be lazy. i'll show you hints only.
1.
molar mass of oxygen = 16x2 = 32 g/mol
molar mass of nitrogen = 14x2 = 28 g/mol
and no. of oxygen molecules = (mass / molar mass) x Avogadro's constant
X = (2/32) x constant
2.
write the formula of these species first.
iron(II) nitrate: Fe(NO3)2
Fe(NO3)2 --->---> Fe(2+) + 2NO3(-)
iron(II) carbonate: FeCO3
FeCO3 --->---> Fe(2+) + CO3(2-)
1 mole of Fe(NO3)2 gives 1 mole of iron(II) ion and 2 mole of nitrate ion;
1 mole of FeCO3 gives 1 mole of iron(II) ion and 1 mole of carbonate ion;
first you have some nitrate -- use his to find no. of iron(II) ions produced by iron(II) nitrate.
then you have total amount of iron(II) ion.
you should be able to calculate amount of iron(II) from the carbonate...
3.
one reagent is in excess, one is limited.
from equation, 1 mole of X reacts with 3 moles of Y. which reagent will be used up first?
also note that 2 moles of XY3 is produced per mole of X. remember to multiply to ratio.
4.
also, find the molar mass of chlorine molecule (Cl2) and sulphur dioxide (SO2).
the calculation is same as question 1.
5.
molar mass of PbSO4 = 207.2 + 32.1 + 16.0x4 = 303.3 g/mol
no. of mole of PbSO4 in the precipitate = mass / molar mass
and, one mole of PbSO4 contains one mole of Pb
so, no. of mole of Pb in the paint sample = ...
and mass = no. of mole x molar mass
mass of Pb in sample = ...
mass percentage of Pb
= 100% x (mass of Pb in sample) / (total mass of the sample )
2011-01-17 17:27:02 補充：
micatkie, please avoid supplying answers directly for ones' homework.
2011-01-17 2:25 am
1)
No. of moles of 2.00 g of O2 = 2.00/(16.0x2) = 1/16 mol
L = Avogadro constant
No. of molecules in 2.00 g of O2:
(1/16)L = x
Hence, L = 16x
No. of moles of 28.0 g of N2= 28.0/(14.0x2) = 1 mol
No. of molecules in 28.0 g of N2 = 1 x L = 16x
= = = = =
2)
Let a mol and b mol be the number of moles of Fe(NO3)2 and FeCO3 respectively.
a mol of Fe(NO3)2 contains a mol of Fe²⁺ ions and 2a mol of NO3⁻ ions.
b mol of FeCO3 contains b mol of Fe²⁺ ions and b mol of CO3²⁻ ions.
No. of NO3⁻ ions: 2a = 1 ⇒ a = 0.5
No. of Fe²⁺ ions: a + b = 1.2 ⇒ 0.5 + b = 1.2 ⇒ b = 0.7
No. of moles of CO3²⁻ ions = b mol = 0.7 mol
= = = = =
3)
Mole ratio in reaction X2 : Y2 = 1 : 3
No. of moles of X2 present = 4 mol
No. of moles of Y2 present = 6 mol
For complete reaction of Y2, no. of moles of X2 needed = 6 * (1/3) = 2 mol
Hence, X2 is in excess, and Y2 completely reacts (limiting reactant).
Mole ratio in reaction Y2 : XY3 = 3 : 2
No. of moles of XY3 formed = 6 x (2/3) = 4 mol
= = = = =
4)
No. of moles of 17.75 g of Cl2 = 17.75/(35.5x2) = 1/4 mol
L = Avogadro constant
No. of molecules in 17.75 g of Cl2:
(1/4)L = x
Hence, L = 4x
No. of moles of 8.00 g of SO2= 8.00/(32.1 + 16x2) = 0.125 mol
No. of molecules in 8.0 g of SO2 = 0.125L = 0.125 * 4x = 0.5x
= = = = =
5)
Molar mass of Pb = 207.2 g/mol
Molar mass of PbSO4 = 207.2 + 32.1 + 16x4 = 303.3 g/mol
Mass of Pb = 0.0849 x (207.2/303.3) = 0.0580 g
% by mass of Pb in the pigment = (0.0580/1.50) x 100% = 3.87%
No. of moles of 2.00 g of O2 = 2.00/(16.0x2) = 1/16 mol
L = Avogadro constant
No. of molecules in 2.00 g of O2:
(1/16)L = x
Hence, L = 16x
No. of moles of 28.0 g of N2= 28.0/(14.0x2) = 1 mol
No. of molecules in 28.0 g of N2 = 1 x L = 16x
= = = = =
2)
Let a mol and b mol be the number of moles of Fe(NO3)2 and FeCO3 respectively.
a mol of Fe(NO3)2 contains a mol of Fe²⁺ ions and 2a mol of NO3⁻ ions.
b mol of FeCO3 contains b mol of Fe²⁺ ions and b mol of CO3²⁻ ions.
No. of NO3⁻ ions: 2a = 1 ⇒ a = 0.5
No. of Fe²⁺ ions: a + b = 1.2 ⇒ 0.5 + b = 1.2 ⇒ b = 0.7
No. of moles of CO3²⁻ ions = b mol = 0.7 mol
= = = = =
3)
Mole ratio in reaction X2 : Y2 = 1 : 3
No. of moles of X2 present = 4 mol
No. of moles of Y2 present = 6 mol
For complete reaction of Y2, no. of moles of X2 needed = 6 * (1/3) = 2 mol
Hence, X2 is in excess, and Y2 completely reacts (limiting reactant).
Mole ratio in reaction Y2 : XY3 = 3 : 2
No. of moles of XY3 formed = 6 x (2/3) = 4 mol
= = = = =
4)
No. of moles of 17.75 g of Cl2 = 17.75/(35.5x2) = 1/4 mol
L = Avogadro constant
No. of molecules in 17.75 g of Cl2:
(1/4)L = x
Hence, L = 4x
No. of moles of 8.00 g of SO2= 8.00/(32.1 + 16x2) = 0.125 mol
No. of molecules in 8.0 g of SO2 = 0.125L = 0.125 * 4x = 0.5x
= = = = =
5)
Molar mass of Pb = 207.2 g/mol
Molar mass of PbSO4 = 207.2 + 32.1 + 16x4 = 303.3 g/mol
Mass of Pb = 0.0849 x (207.2/303.3) = 0.0580 g
% by mass of Pb in the pigment = (0.0580/1.50) x 100% = 3.87%
參考: micatkiela
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