Some Basic Concepts of Chemistry

x      +      y    +   3z     =     xyz₃

1 mole 1 mole 0.05 mole

$?n_x=1$

$n_y=1$

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

$?\frac{0.05}{3}$ mole z gives 1 mole xyz₃

$?$ mass of xyz₃ = n * molecular mass

 = $\frac{0.05}{3}*\left(10+20+3*30\right)a.m.u.$  

= 0.5 * 4 = 2g

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) = $\frac{4}{146}*40=1.095\approx 1.0$

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

$\downarrow$

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 $\frac{4*110}{3*101}$ moles of HNO₃ = 1.45 mole of HNO₃

Hence mass of HNO₃ = mole * mol.wt = 145 * 63 = 91.48 $\approx$ 91.5gm

% of C in organic compound

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

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

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

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

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

= 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 reagent.

$?\frac{0.05}{3}$ mole z gives 1 mole xyz₃

$\therefore$ mass of xyz₃ = n * molecular mass

=  $\frac{0.05}{3}*\left(10+20+3*30\right)a.m.u.$

= 0.5 * 4 = 2g

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 =   $\left(22.5*\frac{756}{210}\right)moles$

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

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

Mass of O₂ liberated =   $15*\frac{756}{210}*44=2376gm$

V = $\frac{}{}\frac{}{}$ $\frac{nRT}{P}=\frac{0.90*0.0821*300}{18*32}*760=29.21\cong 29$

$\frac{0.02858*0.112}{0.5702}=0.0561$

 $C{H}_4+2O_2\to C{O}_2+2H_{2}O$

Mole $\frac{100}{16}\frac{208}{32}$

$\begin{array}{l}\downarrow \downarrow \\ 6.256.5(Here{O}_{2}isthelimitingreagent)\end{array}$

Hence mole of CO₂ formed = $\frac{6.5}{2}$

And wt of CO₂  in gm = $\frac{6.5}{2}*44=143g$

In 4d orbital, n = 4 and $\mathcal{l}=2$

Radial nodes = $n-\mathcal{l}-1$

Radial nodes = 4 – 2 – 1 = 1

And angular nodes,  $\mathcal{l}=2$