Question

calculate the amount of water(H2O) formed during the complete combustion of 96g methane(CH4)(stoichiometric calculation)???

IT'S Urgent
spammers keep away​

Answer 1

Answer:

$\huge\pink{\mid{\underline {\overline{\tt ❥︎A᭄ɴsᴡᴇʀ࿐✪}}\mid}}$

✧══════•❁❀❁•══════✧

CH4+2O2 ⇒CO2+2H2O

1 mol CH4 =2 mol H2O

96 g CH4

.

.

no. of moles = Given mass ÷ Gram molecular mass[GM/GMM]

.

.

GM = 96 g

(C=12,H=1)

GMM=12×1+1×4=16u

.

.

no. of moles⇒96÷16

⇒6 mol

.

.

6 mol CH4 =2×6 mol H2O

=12 mol H2O

.

.

no. of moles⇒GM/GMM

12⇒GM/18

GM =18 ×12

= 216 g

amount of water(H2O)⇒216 g

$\boxed{\fcolorbox{Blue}{Red}{\color{yellow}{hope\:it\:helps\: you}}}$

$\large\boxed{\fcolorbox{green}{blue}{\color{yellow}{MARK\:ME\:AS\: BRAINLIEST}}}$