Question #100b4

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Question #100b4

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Answer 1 · 2017-06-08T02:43:05

Molar masses of

$C_{2} H_{2} \to 2 \times 12 + 2 \times 1 = 26 g / m o l$ $C_2H_2to2xx12+2xx1=26g"/"mol$

$C_{2} H_{4} \to 2 \times 12 + 4 \times 1 = 28 g / m o l$ $C_2H_4to2xx12+4xx1=28g"/"mol$

$C H_{4} \to 1 \times 12 + 4 \times 1 = 16 g / m o l$ $CH_4to1xx12+4xx1=16g"/"mol$

The mole ratio in the mixture

$C_{2} H_{2} : C_{2} H_{4} : C H_{4} = 2 : 1 : 2$ $C_2H_2:C_2H_4:CH_4=2:1:2$

The mass ratio in the mixture

$C_{2} H_{2} : C_{2} H_{4} : C H_{4} = (2 \times 26) : (1 \times 28) : (2 \times 16) = 13 : 7 : 8$ $C_2H_2:C_2H_4:CH_4=(2xx26):(1xx28):(2xx16)=13:7:8$

The masses of the components in the gas mixture of 1g

$C_{2} H_{2} \to \frac{13}{28} g \to \frac{13}{28} \times \frac{1}{26} m o l = \frac{1}{56} m o l$ $C_2H_2to13/28g-> 13/28xx1/26mol=1/56mol$

$C_{2} H_{4} \to \frac{7}{28} g \to \frac{7}{28} \times \frac{1}{28} m o l = \frac{1}{112} m o l$ $C_2H_4to7/28g->7/28xx1/28mol=1/112mol$

$C H_{4} \to \frac{8}{28} g \to \frac{8}{28} \times \frac{1}{16} m o l = \frac{1}{56} m o l$ $CH_4to8/28g->8/28xx1/16mol=1/56mol$

Balanced equation of the reactions

$C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l)$ $C_2H_2(g)+5/2O_2(g)->2CO_2(g)+H_2O(l)$

$C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l)$ $C_2H_4(g)+3O_2(g)->2CO_2(g)+2H_2O(l)$

$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ $CH_4(g)+2O_2(g)->CO_2(g)+2H_2O(l)$

So the number of moles of $O_{2} (g)$ $color(red)(O_2(g))$ required for complete burning of component gases are

$C_{2} H_{2} \to \frac{1}{56} \times \frac{5}{2} = \frac{5}{112} m o l$ $C_2H_2to1/56xx5/2=5/112mol$

$C_{2} H_{4} \to \frac{1}{112} \times 3 = \frac{3}{112} m o l$ $C_2H_4to1/112xx3=3/112mol$

$C H_{4} \to \frac{1}{56} \times 2 = \frac{4}{112} m o l$ $CH_4to1/56xx2=4/112mol$

So total amount of $O_{2} (g)$ $color(red)(O_2(g))$ required for complete burning of 1 g gas mixture is

$O_{2} \to \frac{5 + 3 + 4}{112} m o l = \frac{12}{112} m o l$ $O_2->(5+3+4)/112mol=12/112mol$

As the ratio of $O_{2} : N_{2}$ $O_2 :N_2$ by volume in air is $(20 %) : (80 %) = 1 : 4$ $(20%):(80%)=1:4$

The mole ratio of $O_{2} : N_{2}$ $O_2 :N_2$ in air will be $= 1 : 4$ $=1:4$

So in air $\frac{12}{112} m o l$ $12/112mol$ $O_{2}$ $O_2$ will be with $\frac{12}{112} \times 4 m o l$ $12/112xx4mol$ $N_{2}$ $N_2$

Hence total mass of air required for complete burning is

$\frac{12}{112} m o l$ $12/112mol$ $O_{2} +$ $O_2+$ $\frac{12}{112} \times 4 m o l$ $12/112xx4mol$ $N_{2}$ $N_2$

$= \frac{12 \times 32 + 48 \times 28}{112} g = \frac{1728}{112} g \to$ $=(12xx32+48xx28)/112g=1728/112g->$ option A

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Question #100b4

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