Chemistry question is hard for me to answer.
chlorine dioxide, clo2, has been used as a disinfectant in air conditioning systems. it reacts with water according to the equation:
6c102 +3h20 5hcio3 + hci
if 119.3 g of clo2 is mixed with 48.0 g of h20, how many grams of the excess reactant remain when the reaction is complete?

0.161 grams

Explanation:

You first need to find the weight that CLO2 uses first.

So to get that answer, you would do, 35.45(chlorine) + 16.00(oxygen)+ 16.00(oxygen) and that equals 77.45.

So to calculate the number of moles, you would do 119.3/77.45 and that would give you 1.540.

Now you do the same thing with water.  1.01 (water) + 1.01(water) + 16.00(oxygen) which is 18.02.

So now you divide 48.0/18.02 and that is 2.664

Looking at the equation that tells us that for 1.833 moles of ClO2 needs 0.917 moles of H2O.

That tells us that water is also the excess.

So when you do the calculation for water, the amount of water left would be 2.664-0.917 and that equals to 2.905 g.  

Then you would multiply 2.905 by 18.02, and your final answer is:

0.161 g

(Hopefully you find this helpful!!)

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fe 55.85 g/mol and h2o 18.02 g/mol.