In which reaction does the smallest percentage change in volume occur?
a) C3H8 (g) + 5O2(g) ---> 3CO2(g) + 4H2O(l)
b) 4NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
c) CH4(g)+ 2O2(g) ----> CO2(g) + 2H2O(l)
d) 2H2S(g) + SO2 (g) ----> 3S(s) + 2H2O(l)
The correct answer is A, but how do i do this?
sorry i submitted the question twice i think .
To determine which reaction has the smallest percentage change in volume, we need to compare the stoichiometric coefficients of the reactants and products.
Consider the balanced equation for each reaction.
a) C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(l)
b) 4NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
c) CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(l)
d) 2H2S(g) + SO2(g) ----> 3S(s) + 2H2O(l)
The stoichiometric coefficients represent the ratio of moles of substances involved in the reaction.
In reaction a, the stoichiometric coefficient of CO2 is 3, while in reactions b and c, the stoichiometric coefficient of N2 is 2.
To determine the percentage change in volume:
1. Calculate the moles of the reactants and products in each reaction using stoichiometric coefficients.
2. Determine the volume ratios based on the stoichiometric coefficients of the gases.
3. Compare the volume ratios of the gases involved in each reaction.
4. The reaction with the smallest change in volume will have the smallest percentage change.
Let's calculate the moles and volume ratios for each reaction:
a) C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(l)
Here, CO2 has a stoichiometric coefficient of 3. Let's assume all gases have equal volumes initially:
Initial volume ratio: 1:5 (C3H8:O2)
Final volume ratio: 3:4 (CO2:H2O)
b) 4NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
Here, N2 has a stoichiometric coefficient of 2. Let's assume all gases have equal volumes initially:
Initial volume ratio: 4:3 (NH3:O2)
Final volume ratio: 2:6 (N2:H2O)
c) CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(l)
Here, CO2 has a stoichiometric coefficient of 1. Let's assume all gases have equal volumes initially:
Initial volume ratio: 1:2 (CH4:O2)
Final volume ratio: 1:2 (CO2:H2O)
d) 2H2S(g) + SO2(g) ----> 3S(s) + 2H2O(l)
In this case, the products are solids, so we do not consider their volume.
Let's assume all gases have equal volumes initially:
Initial volume ratio: 2:1 (H2S:SO2)
Comparing the volume ratios for each reaction:
a) Initial volume ratio: 1:5, Final volume ratio: 3:4
b) Initial volume ratio: 4:3, Final volume ratio: 2:6
c) Initial volume ratio: 1:2, Final volume ratio: 1:2
d) Initial volume ratio: 2:1
From the volume ratios, it is clear that reaction a has the smallest change in volume, as the final volume ratio is closest to the initial volume ratio. Therefore, the correct answer is A.