Question

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

Answer 1

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 = $0.998g\times 2.07\% =0.998g\times \frac{2.07}{100}=0.0206g$

Now we have to calculate the moles of Ni.

$\text{Moles of }Ni=\frac{\text{Given mass }Ni}{\text{Molar mass }Ni}=\frac{0.0206g}{58.7g/mol}=0.0003509mol$

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.

$\text{ Mass of dimethylglyoxime}=\text{ Moles of dimethylglyoxime}\times \text{ Molar mass of dimethylglyoxime}$

Molar mass of dimethylglyoxime = 116.12 g/mole

$\text{ Mass of dimethylglyoxime}=(0.0010527moles)\times (116.12g/mole)=0.1222g$

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 $\frac{100}{2.15}\times 0.1222=5.684g$ of solution

Now we have to calculate the volume of solution.

$Density=\frac{Mass}{Volume}$

$0.790g/mL=\frac{5.684g}{Volume}$

Volume of solution = 7.195 mL

Therefore, the volume of solution is, 7.195 mL