91Ó°ÊÓ

Let us calculate the moles nitrogen present in the fertilizer

Given:

$\text{Molar mass of}$${\text{N}}_2$=$28.02\text{g}$

$28.02{\text{g ofN}}_2\text{contains = 1 mole}$

$\text{Mass percentage of}$${\text{N}}_2\text{=30\%}$

${\text{30\% ofN}}_{\text{2}}{\text{means, 0.30 grams ofN}}_{\text{2}}$

Therefore,

${\text{0.30g ofN}}_{\text{2}}\text{contains =}\frac{1×0.30}{{\displaystyle \text{28.02}}}$

$=0.0107\text{moles}$

Let us calculate the moles phosphorous present in the fertilizer

Given:

${\text{Molar mass ofP}}_2{\text{O}}_5=141.94\text{g/mol}$

${\text{Mass percentage ofP}}_2{\text{O}}_5\text{= 10\%}$

$141.94{\text{g ofP}}_2{\text{O}}_5\text{contains = 1 mole}$

$$\begin{gathered} 0.10{\text{g ofP}}_2{\text{O}}_5\text{contains =}\frac{1\text{mole}×0.10\text{g}}{141.94\text{g}} \\ =0.0007\text{moles} \end{gathered}$$

Let us calculate the moles potassium present in fertilizer

${\text{Molar mass ofK}}_2\text{O}=94.2\text{g/mol}$

$94.2{\text{g ofK}}_2\text{O contains = 1 mole}$$94.2{\text{g ofK}}_2\text{O contains = 1 mole}$

$$\begin{gathered} \text{10}\%{\text{ofK}}_2{\text{O means, 0.10 grams ofK}}_2\text{O} \\ 0.10{\text{g ofK}}_2\text{O contains =}\frac{1\text{moles}×0.10\text{g}}{94.2\text{g}} \\ =0.0011\text{moles} \end{gathered}$$

Molar ratio of N/P/K = $\text{0.0107/0.0007/0.0011}$

Simple ratio: $$\begin{gathered} \frac{\text{0.0107}}{\text{0.0011}}:\frac{\text{0.0007}}{\text{0.0011}}:\frac{\text{0.0011}}{\text{0.0011}} \\ =10/0.66/1.0 \end{gathered}$$

Result:

The molar ratio of N/P/K in a fertilizer$=10/0.66/1.0$