Salam and good day to all,
Today, we discussed lengthy about stoichiometry/quantitative analysis. Among the concepts that we covered today are mol of solution, atom, ion, molecule, limiting reagent and gas law. I found that the students have difficulty in changing chemical statement to mathematical equation. These are some of the questions that we went thru;
1. Consider the following equation:
N2(g) + 3H2(g) gives 2NH3(g)
If the reaction is made to go to completion, what volume of ammonia (in dm3) can be prepared from 25 dm3 of nitogen and
60 dm3 of hydrogen? All volumes are measured at the same temperature and pressure.
A. 40
B. 50
C. 85
D. 120
Solution : If 1 mol of N2 used 25dm3 ,3 mol of H2 needed 75 dm3. But given only 60dm3. it means that H2 is not enough. It is the limiting reagent. So mol of ammonia depends on mol of H2. Therefore mol of NH3 is 60 times 2 divided by 3. The answer is 40 which is A.
2. What is the empirical formula of a compound containing 50% by mass of element X (R.A.M. = 20) and 50% by mass of element Y (R.A.M. = 25)?
A. XY
B. X3Y2
C. X4Y5
D. X5Y4
Solution : Get the no. of mole of X and Y.
For X; mol = 50/20 = 2.5
Y; mol = 50/25 = 2
So ratio between X : Y
= 2.5 : 2
Times both no. with closest whole no. so that you don’t have decimal no. for X. Therefore times both of them with 2.
So you will get for X, 2.5 x 2 = 5
while for Y, 2 x 2 = 4
Therefore empirical formula is X5Y4. So the answer is D.
3. What volume of 0.500moldm-3 sulfuric acid solution is required to react completely with 10.0 g of calcium carbonate according to the equation below?
CaCO3(s) + H2SO4(aq) → CaSO4(aq) + H2O(l) + CO2(g)
A. 100cm3
B. 200cm3
C. 300cm3
D. 400cm3
Solution : Calculate the no. of mole of CaCO3 first since it is the known substance.
RMM of CaCO3 = 40 + 12 * 16(3) = 100
no. of mole of CaCO3 = 10.0/RMM CaCO3
= 10.0/100
= 0.100 mole
From equation 1 mole of CaCO3 react with 1 mole of H2SO4, therefore
0.100 mole of CaCO3 recat with 0.100 mol of H2SO4.
To find mole of H2SO4,
use mv/1000 for mole of solution,
so mv/1000 = 0.100
molarity of H2SO4 is known which is 0.500moldm-3, therefore
(0.500 dm-3 x V)/1000 = 0.100
V = (0.100 x1000)/0.5
V = 1000/5
V = 200 cm3
S0, the answer is B.
Question 4 (Newly added).
A 4.0 gram sample of sodium hydroxide, NaOH is dissolved in water and made up to 500 cm3 of solution. What is the concentration of resulting solution?
A. 0.1 moldm-3
B. 0.2 moldm-3
C. 0.5 moldm-3
D. 1.0 moldm-3.
Solution :
First get the no. of mol of sodium hydroxide. To get no. of mol of sodium hydroxide, we need it’s relative molecular mass.
So relative molecular mass of NaOH = 23 + 16 + 1
= 40
then no. of mol of NaOH = mass/relative molecular mass
= 4/40
= 0.1 mol
concentration = mol of NaOH/ volume of solution in dm3
= 0.1 mol /0.5 dm3
= 0.2 moldm-3.
Therefore the answer is B.
Have fun learning chemistry…
