How much 6.0 M HNO3 is needed to neutralize 39 mL of 2.0 M KOH?

2 answers:

7 0

san4es73 [151]3 years ago

4 0

Answer : The volume of $HNO_3$ needed are, 0.013 L

Explanation :

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of $HNO_3$ = 6.0 M = 6.0 mole/L

$M_2$ = molarity of $KOH$ = 2.0 M = 2.0 mole/L

$V_1$ = volume of $HNO_3$ = ?

$V_2$ = volume of $KOH$ = 39 ml = 0.039 L

conversion : 1 L = 1000 ml

Now put all the given values in the above formula, we get the volume of $HNO_3$

$6.0mole/L\times V_1=2.0mole/L\times 0.039L$

$V_1=0.013L$

Therefore, the volume of $HNO_3$ needed are, 0.013 L

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The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen

tino4ka555 [31]

Answer: The mass percentage of $Tl_2SO_4$ is 5.86%

Explanation:

To calculate the mass percentage of $Tl_2SO_4$ in the sample it is necessary to know the mass of the solute ( $Tl_2SO_4$ in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the $TlI$ precipitate. We can establish a relation between the mass of $TlI$ and $Tl_2SO_4$ using the stoichiometry of the compounds:

$moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.$

Since for every mole of Tl in $TlI$ there are two moles of Tl in $Tl_2SO_4$ , we have:

$moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol$

Using the molar mass of $Tl_2SO_4$ we have:

$mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g$

Finally, we can use the mass percentage formula:

$mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%$