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3 years ago

What is the percent of Li in Li3N ? Please!!

Chemistry

2 answers:

lisabon 2012 [21]3 years ago

6 0

I think it maybe 59.7852%

rjkz [21]3 years ago

3 0

Answer: Li percentage in $Li_3N$ is 59.8%.

Explanation: The question asks to calculate the mass percentage of Li in $Li_3N$ .

The formula used for mass percentage of an element in a compound is:

mass percentage = $(\frac{element mass}{compound mass})*100$

Atomic mass of Li is 6.94 and atomic mass of N is 14.0.

There are three Li in the given compound. So, the mass of Li in the compound is = 3(6.94) = 20.82

Mass of compound = 20.82 + 14.0 = 34.82

Mass percentage of Li = $(\frac{20.82}{34.82})*100$ = 59.8%

Hence, the percentage of Li in $Li_3N$ is 59.8%.

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$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

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Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

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