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?
look on the left side, and the right side, for the coefficents of the gaseous products.
a. 6>3 ratio .5
b. 7>2 ratio .286
c. 3>1 ratio .33
d. 3>0 ratio 0.0
so which changed volumes the least?
Well, determining which reaction has the smallest percentage change in volume can be a bit tricky. But don't worry, I'm here to clown around and help you figure it out!
First, let's look at each reaction and see if any of them involve gases going into liquid or solid forms. In reactions b, c, and d, we have a liquid or solid product. However, in reaction a, we only have gases as products, which means there is no change from a gaseous state to a liquid or solid state.
Now, when gases react and form other gases, the overall volume might change. To determine the percentage change, you need to find the difference in the number of moles of gas on each side of the reaction equation and then divide it by the initial number of moles.
In reaction a, we start with propane (C3H8) and oxygen (O2) as gases and form carbon dioxide (CO2) and water (H2O) as gases and a liquid, respectively. Here, we have 3 moles of CO2 and 4 moles of H2O on the product side. On the reactant side, we have 1 mole of C3H8 and 5 moles of O2. Since the stoichiometric coefficients of the gases are similar on both sides (3 vs. 1 for CO2 and 4 vs. 5 for H2O), the change in volume is relatively small, resulting in the smallest percentage change in volume. Hence, option A is the correct answer.
Remember, it's not the size of the reaction, but the size of the volume change that counts!
To determine the reaction with the smallest percentage change in volume, you need to analyze the stoichiometry of the reactions and calculate the molar volume change.
First, we need to determine the change in the number of moles for each reaction. This can be done by comparing the stoichiometric coefficients of the reactants and products.
a) C3H8 (g) + 5O2(g) ---> 3CO2(g) + 4H2O(l)
In this reaction, 1 mole of C3H8 reacts with 5 moles of O2, producing 3 moles of CO2 and 4 moles of H2O. Therefore, the change in the number of moles is 3 - 1 = 2 moles.
b) 4NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
In this reaction, 4 moles of NH3 reacts with 3 moles of O2, producing 2 moles of N2 and 6 moles of H2O. Therefore, the change in the number of moles is 2 - 4 = -2 moles.
c) CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(l)
In this reaction, 1 mole of CH4 reacts with 2 moles of O2, producing 1 mole of CO2 and 2 moles of H2O. Therefore, the change in the number of moles is 1-1 = 0 moles.
d) 2H2S(g) + SO2 (g) ----> 3S(s) + 2H2O(l)
In this reaction, 2 moles of H2S react with 1 mole of SO2, producing 3 moles of S and 2 moles of H2O. Therefore, the change in the number of moles is 3 - 2 = 1 mole.
Next, we need to consider the molar volumes of the gases. The molar volume is the volume occupied by 1 mole of gas at a specific temperature and pressure (usually measured in liters).
Since the volume change is directly proportional to the change in moles, we can compare the magnitude of molar volume change using stoichiometry.
By evaluating the changes in the number of moles, we can see that option c) CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(l) has the smallest change in moles (0). Therefore, it will have the smallest percentage change in volume.
To determine which reaction produces the smallest percentage change in volume, we can use the concept of stoichiometry. By comparing the stoichiometric coefficients of the reactants and products, we can analyze the relationship between the number of moles of each species involved in the reaction.
In this case, we are comparing the change in volume, so we need to consider the relationship between gases. Generally, the ideal gas law states that the volume of a gas is directly proportional to the number of moles of that gas.
Let's assess each reaction in turn:
a) C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(l)
In this reaction, the molar ratio of the reactant (C3H8) to the products (CO2) is 1:3. This means that for every mole of C3H8 consumed, three moles of CO2 are produced. Since CO2 is a gas, the volume will increase with this reaction.
b) 4NH3(g) + 3O2(g) ----> 2N2(g) + 6H2O(l)
In this reaction, the molar ratio of the reactant (NH3) to the products (N2) is 4:2 or 2:1. This means that for every two moles of NH3 consumed, one mole of N2 is produced. Since N2 is a gas, the volume will decrease with this reaction.
c) CH4(g)+ 2O2(g) ----> CO2(g) + 2H2O(l)
In this reaction, the molar ratio of the reactant (CH4) to the products (CO2) is 1:1. This means that for every mole of CH4 consumed, one mole of CO2 is produced. The volume will not change significantly since the moles of gas are equal on both sides of the reaction.
d) 2H2S(g) + SO2(g) ----> 3S(s) + 2H2O(l)
In this reaction, the molar ratio of the reactant (H2S) to the products (S) is 2:3. This means that for every two moles of H2S consumed, three moles of S are produced. Since S is a solid, it does not occupy a significant volume, resulting in a relatively small change in volume.
Comparing the reactions, we can see that the reaction with the smallest percentage change in volume is c) CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l).