2HCr4- + 3HS03- + 5H+ -> 2Cr3+ + 3SO4^2- + 5H20.
At t=0, 0.0249 mols of HCr04- are loaded in to 250ml of solution. After 1/10 of an hour, 0.00852 mol of HCrO4- remain. Assuming the solution volume remained the same, how many grams of water were formed per second during this interval of time on average?
TIA!
according to the balanced equation, for every 4 moles of HCrO4, you made 5 moles of water.
You know the amount of HCrO4 used (initial-final), so take 5/4 of that to get moles of water, then divide by time to get average rate. (1/10 of a hour is 360 second)
Another question is when you invert it and times everything by negative 3 it would make the rates positive on the reactants side and negative on the products side. When you get the rate for that, do you make it positive or negative and why?
To calculate the grams of water formed per second during this interval of time, we need to determine the change in the number of moles of water formed and then convert it to grams.
First, let's calculate the initial number of moles of HCrO4-:
Initial moles of HCrO4- = 0.0249 mol
Next, let's calculate the final number of moles of HCrO4-:
Final moles of HCrO4- = 0.00852 mol
Now, we can calculate the change in moles of HCrO4- during this interval of time:
Change in moles of HCrO4- = Initial moles of HCrO4- - Final moles of HCrO4-
Change in moles of HCrO4- = 0.0249 mol - 0.00852 mol = 0.01638 mol
Since the balanced chemical equation tells us that 5 moles of water are formed for every 2 moles of HCrO4-, we can determine the moles of water formed:
Moles of water formed = (5/2) x Change in moles of HCrO4-
Moles of water formed = (5/2) x 0.01638 mol = 0.04095 mol
Now, we need to convert the moles of water to grams:
Molar mass of water (H2O) = 18.015 g/mol
Grams of water formed = Moles of water formed x Molar mass of water
Grams of water formed = 0.04095 mol x 18.015 g/mol ≈ 0.738 g
Since the interval of time is 1/10 of an hour, which is equal to 6 seconds, we can calculate the grams of water formed per second:
Grams of water formed per second = Grams of water formed / Interval of time
Grams of water formed per second = 0.738 g / 6 s ≈ 0.123 g/s
Therefore, approximately 0.123 grams of water were formed per second on average during this interval of time.
To determine the grams of water formed per second during this interval of time, we need to calculate the change in the number of moles of water over the time interval and then convert it to grams.
Given reaction:
2HCrO4- + 3HSO3- + 5H+ -> 2Cr3+ + 3SO4^2- + 5H2O.
From the balanced equation, we can see that for every 5 moles of HCrO4-, 5 moles of water are produced.
Step 1: Calculate the moles of HCrO4- consumed:
At t=0, there are 0.0249 moles of HCrO4-.
After 1/10 of an hour, there are 0.00852 moles of HCrO4- remaining.
Therefore, the number of moles of HCrO4- consumed = 0.0249 - 0.00852 = 0.01638 moles.
Step 2: Calculate the moles of water formed:
Since the ratio of HCrO4- to water is 1:1, the number of moles of water formed is also equal to 0.01638 moles.
Step 3: Convert moles of water to grams:
To convert moles of water to grams, we need to use the molar mass of water.
Molar mass of water (H2O) = 2(1.008 g/mol) + 16.00 g/mol = 18.02 g/mol.
Mass of water formed = moles of water × molar mass of water
Mass of water formed = 0.01638 moles × 18.02 g/mol = 0.295 g.
Step 4: Calculate the average grams of water formed per second:
Since the time interval is 1/10 of an hour, or 6 seconds (1/10 hour × 60 minutes/hour × 60 seconds/minute = 6 seconds),
Average grams of water formed per second = mass of water formed / time interval
Average grams of water formed per second = 0.295 g / 6 s = 0.049 g/s.
Therefore, the average grams of water formed per second during this interval of time is 0.049 grams.