Asked by Tri

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!
Answered by bobpursley
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)
Answered by Tri
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?
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