How many milliliters of 2.15 wt% dimethylglyoxime solution should be used to provide a 50.0% excess for Reac- tion 7-2 with 0.998 4 g of steel containing 2.07 wt% Ni

This is an incomplete question, here is a complete question.

How many milliliters of 2.15 wt% dimethylglyoxime solution should be used to provide a 50.0% excess for Reaction 7-2 with 0.998 4 g of steel containing 2.07 wt% Ni?  Assume that the density of the dimethylglyoxime solution is 0.790 g/mL.

Answer : The volume of solution is, 7.195 mL

Explanation :

First we have to calculate the mass of Ni.

Mass of Ni =

Now we have to calculate the moles of Ni.

Now we have to calculate the moles of dimethylglyoxime.

As we know that 1 mole of Ni requires 2 moles of dimethylglyoxime.

So, 0.0003509 mole of  Ni requires (0.0003509×2=0.0007018) moles of dimethylglyoxime.

For 50% excess reaction number of moles of dimethylglyoxime = 0.0007018 × 1.5 = 0.0010527 g

Now we have to calculate the mass of dimethylglyoxime.

Molar mass of dimethylglyoxime = 116.12 g/mole

Now we have to calculate the weight of solution.

As, 2.15 g of dimethylglyoxime present in 100 g of solution

So, 0.1222 g of dimethylglyoxime present in of solution

Now we have to calculate the volume of solution.

Volume of solution = 7.195 mL

Therefore, the volume of solution is, 7.195 mL

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

65uuyhjghj8ykitdyidii