Respuesta :
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 = [tex]0.998g\times 2.07\% =0.998g\times \frac{2.07}{100}=0.0206g[/tex]
Now we have to calculate the moles of Ni.
[tex]\text{Moles of }Ni=\frac{\text{Given mass }Ni}{\text{Molar mass }Ni}=\frac{0.0206g}{58.7g/mol}=0.0003509mol[/tex]
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.
[tex]\text{ Mass of dimethylglyoxime}=\text{ Moles of dimethylglyoxime}\times \text{ Molar mass of dimethylglyoxime}[/tex]
Molar mass of dimethylglyoxime = 116.12 g/mole
[tex]\text{ Mass of dimethylglyoxime}=(0.0010527moles)\times (116.12g/mole)=0.1222g[/tex]
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 [tex]\frac{100}{2.15}\times 0.1222=5.684g[/tex] of solution
Now we have to calculate the volume of solution.
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]0.790g/mL=\frac{5.684g}{Volume}[/tex]
Volume of solution = 7.195 mL
Therefore, the volume of solution is, 7.195 mL
