Some Basic Concepts of Chemistry

Total wt = 4gm = W_NaOH + $W_{N{a}_{2}C{O}_3}$  

Let us suppose moles are ‘m’ for each.

4 = 40m + 106m

$\therefore m=\left(\frac{4}{146}\right)$ moles of NaOH and Na₂CO₃ each .

$\therefore$ Mass of NaOH (x gm) = $\frac{4}{146}\times 40=1.095\approx 1.0$

$4HNO{I}_{3\left(\mathcal{l}\right)}+3KC{l}_{\left(s\right)}\underset{}{\overset{}{\to }}C{l}_{2\left(g\right)}+NOC{l}_{\left(g\right)}+2H_{2}O\left(g\right)+3KN{O}_3\left(g\right)$

$\downarrow$

4 moles of HNO₃ produced 3 mol of KNO₃

Here mole of produced KNO₃ = $\frac{110}{101}$

If 3 mol of KNO₃ produced by 4 moles of HNO₃

$\therefore$ 1 mole of KNO₃ produced by $\frac{4}{3}$ moles of HNO₃

and $\frac{110}{101}$ mole of KNO₃ produced by $\frac{4\times 110}{3\times 101}$ moles of HNO₃ = 1.45 mole of HNO₃

Hence mass of HNO₃ = mole × mol.wt = 145 ×

% of C in organic compound

$=\frac{12}{14}\times \frac{WC{O}_2}{W.O.C.}\times 100$

$=\frac{952.56}{21.648}=44\mathrm{\%}$

$\mathrm{\%}ofH=\frac{2}{18}\times \frac{W{H}_{2}O}{W.O.C.}\times 100$

= $\frac{20}{18}\times \frac{0.4428}{0.492}\times 100$ $\frac{}{}\frac{}{}$

= $\frac{88.56}{8.856}=10\mathrm{\%}$ $\frac{}{}$

= 100 – 54 = 46%

x      +      y    +   3z     =     xyz₃

1 mole 1 mole 0.05 mole

$\therefore n_x=1$          

  $n_y=1$              

$n_z=\frac{0.05}{3}=0.0167$ here z is limiting reag

Mass of C₁₅H₃₀ = volume × Density

= 1000 $m\mathcal{l}$  × 0.756 gm/ $m\mathcal{l}$  

C₁₅H₃₀ + 22.5 O₂   $\stackrel{}{\to }$ 15CO₂ + 15H₂O

No. of moles of C₁₅H₃₀ = $\frac{756}{210}$ moles

No. of moles of O₂ required =   $\left(22.5\times \frac{756}{210}\right)moles$

Mass of O₂ required = 22.5 ×   $\frac{756}{210}\times 32=2592gm$

No. of moles of CO₂ liberated = 15 ×   $\left(\frac{756}{210}\right)$ moles

Mass of O₂ liberated = Views|Posted a year ago