xg of methane was burnt completely in the presence of oxygen. The liberated gases were passed into a solution containing 370g of Ca ( OH )_2. The weight of white precipitate obtained was 500 . What i…

Question

xg of methane was burnt completely in the presence of oxygen. The liberated gases were passed into a solution containing 370g of Ca ( OH )_2. The weight of white precipitate obtained was 500 . What is the value of x (in g )? (Given : C=12 ;H=1 ; Ca=40 ;O=16u )

Accepted Answer

Correct answer: B. (b) CH_4+2O_2 CO_2+2H_2O Thus, 1 mole of CH_4 gives 1 mole of CO_2 gas. CO_2 gas combines with Ca ( OH )_2 as : Ca ( OH )_2+ CO_2 CaCO CO_3 ↓+H_2O Thus, 1 mole of Ca ( OH )_2 gives 1 mole of CaCO_3 precipitate or 74 gCa ( OH )_2 gives 100 g of CaCO_3 precipitate. Therefore, 500 g of precipitate would be obtained by : 74 5=370 g or 5 moles of Ca ( OH )_2. Since products are formed as per the stoichiometric coefficients, there would be 5 moles of CO_2 released in the first reaction. 5 moles of CO_2 would be released by 5 moles or 5 16=80 g of CH_4. Thus, x=80.

$xg of methane was burnt completely in the presence of oxygen. The liberated gases were passed into a solution containing 370 g of Ca (OH)_{2} . The weight of white precipitate obtained was 500 . What is the value of x (in g)? (Given : C = 12; H = 1; Ca = 40; O = 16 u)$

Solution

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