Question

the 10 gram sample of natural gas containing ch4 and C2 H4 was burnt in excess of oxygen to give 29 gram of seo2 and some water how many gram of water are formed.

Answer 1

Combustion of methane is

$CH_4  + O_2   ->   CO_2  + 2 H_2O$

$m = x g$

$n = x /16$

As, x /16 mol of CH4 produces x /16 mole CO2

Combustion of ehtene is

$C_2H_4  +  3 O_2 >   2 CO_2 +  2 H_2O$

$m = (10 – x) g$

$n = (10 – x) /28$

As, 1 mol of C2H4 produces 2 mol CO2

$(10 – x) /28 mol C2H4 produce =[2 *  (10 – x)] /28 mol CO2$

As, 29 g Co2 is produced during the reaction,

$Mole of CO2 = m /M = 29 /44 = 0.65 mol$

So, mol of CO2 from both the reaction combines to produce 0.65 mol

$x /16 +  (10 – x) /14 = 0.65$

$x /8 + (10-x) /7 = 1.3$

$(7 x + 80 – 8 x) = 72.8$

$X = 7.2 g$

$10 – x = 2.8 g$

Thus, 7.2 g methane and 2.8 g ethene is reacted with oxygen

From 1st reaction 1 mol methane produces 2 mol water  

Mol of methane = n = x /16

$n = 7.2 / 16 = 0.45 mol$

Mol of water = 2 * 0.45 = 0.9 mol

m /M = n  

.m /18 = 0.9

. m = 16.2 g [tex][/tex]

Thus, 16.2 g of water is produced.

Answer 2

Answer:

Explanation:

CH4 +2O2------>CO2+2H2O

C2H4+3O2------>2CO2+2H2O

Let CH4=X

•Mole CH4 =X/16

Now 1 mole CH4 give 1 mole CO2

Hence,X/16 mole give Mole of CO2 =X/16

||ly (10-X)/28 mole of C2H4 gives 10-X/14 mole of CO2

Hence,Total mole of CO2 =14x+16 (10-x)/224

NOW,we are given that 29g CO2 iso formed find mole put equal to above eq

Get X findex mole of CH4 and C2H4 find find mole of water and weight of h20