Consider the combustion of ethane:
2C2H6(g)+7O2(g)-->4CO2(g)+6H2O(g)
If the ethane is burning at the rate of 0.7 mol/L × s, at what rates are and being produced?
yes that's what he was asking, thank you!
Well, well, well, looks like we have a little combustion party going on here! Let's see what's cooking, shall we?
According to the balanced equation, for every 2 moles of ethane burned, we get 4 moles of carbon dioxide and 6 moles of water. So, the molar ratio is 2:4:6.
Now, if the ethane is burning at a rate of 0.7 mol/L × s, we just need to do a little math to find out the rate of carbon dioxide and water being produced.
For every 2 moles of ethane burned, we get 4 moles of carbon dioxide. Therefore, the rate of carbon dioxide being produced is (0.7 mol/L × s) × (4 mol ethane / 2 mol carbon dioxide) = 1.4 mol/L × s.
Similarly, for every 2 moles of ethane burned, we get 6 moles of water. So, the rate of water being produced is (0.7 mol/L × s) × (6 mol ethane / 2 mol water) = 2.1 mol/L × s.
So, to summarize, we have the ethane burning at the rate of 0.7 mol/L × s, carbon dioxide being produced at the rate of 1.4 mol/L × s, and water being produced at the rate of 2.1 mol/L × s.
It's quite a fiery dance party indeed! Keep the fire jokes away, though. Safety first!
In the given chemical equation, the stoichiometric coefficients represent the molar ratios between the reactants and products. Using these coefficients, we can determine the rates at which each substance is being produced or consumed.
Given:
Rate of ethane combustion = 0.7 mol/L × s
From the balanced equation, we see that the stoichiometric ratio between ethane (C2H6) and carbon dioxide (CO2) is 2:4. Therefore, for every 2 moles of ethane consumed, 4 moles of carbon dioxide are produced.
So, the rate at which carbon dioxide is being produced can be calculated as follows:
Rate of CO2 production = (4/2) × (Rate of ethane combustion)
= 2 × 0.7 mol/L × s
= 1.4 mol/L × s
Now, let's consider the ratio between ethane and water (H2O) in the balanced equation, which is 2:6. For every 2 moles of ethane consumed, 6 moles of water are produced.
Thus, the rate at which water is being produced can be calculated as:
Rate of H2O production = (6/2) × (Rate of ethane combustion)
= 3 × 0.7 mol/L × s
= 2.1 mol/L × s
Therefore, the rate of carbon dioxide production is 1.4 mol/L × s, and the rate of water production is 2.1 mol/L × s.
To determine the rates at which carbon dioxide (CO2) and water (H2O) are being produced, we need to use the stoichiometry of the balanced chemical equation:
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
The balanced equation shows that 2 moles of ethane (C2H6) react to produce 4 moles of carbon dioxide (CO2) and 6 moles of water (H2O).
The given rate is 0.7 mol/L · s, which represents the rate of ethane consumption.
To find the rates of carbon dioxide and water production, we will use the stoichiometric coefficients from the balanced equation.
Rate of CO2 production:
Looking at the equation, 2 moles of ethane produce 4 moles of CO2. Therefore, the ratio of the rates is 4:2 or 2:1.
Rate of CO2 production = 2/1 * Rate of ethane consumption = 2/1 * 0.7 mol/L · s = 1.4 mol/L · s
Therefore, the rate of carbon dioxide production is 1.4 mol/L · s.
Rate of H2O production:
Similarly, 2 moles of ethane produce 6 moles of H2O. The ratio of the rates is 6:2 or 3:1.
Rate of H2O production = 3/1 * Rate of ethane consumption = 3/1 * 0.7 mol/L · s = 2.1 mol/L · s
Therefore, the rate of water production is 2.1 mol/L · s.
at what rates ???? are being produced..
the mole ratios are handy
well, you get twice as much CO2 as ethane, mole wise. Now at the same temp, pressure, this means twice as much volume also
so, lets just consider the moles first.
ethane is being consumed at .7mol/sec, which means CO2 is being produced at 1.4 moles per sec, and water is being produced at 2.1 moles/second.
Now consider volume. All are gases, at the same temp pressure, moles occupy the same volume, so volume wise..
for each liter of ethane, we get two liters of CO2, and three liters of H2O.
so in the rate you specified, in moles per liter per second...
if ethane is being consumed at .7 mol/liter.sec, then CO2 is being produced at the rate of 1.4moles/2 liters (or .7 moles/liter), and likewise, water is being produced at .7mol/liter.sec
I am not certain your instructor intended to ask for this result.