Chemistry Daily Challenge 28-July-2015

$\color{#3D99F6}{\text{1) First write chemical equation.}}$

$\ce{(CH4)_x(H2O)_y + a O2 -> bCO2 + xH2O}$

$\color{#3D99F6}{\text{2) Write balance equation for C,O and H.}}$

Carbon balance equation : $\quad x=b$

Hydrogen balance equation: $\quad 4x+2y=2c$

Oxygen balance equation : $\quad y+2a=2b+c$

$\color{#3D99F6}{\text{3) Find number of mole of methane hydrate,water and carbon di oxide}}$

Number of mole of methane hydate= $\frac{100}{2835.18}=0.03527 \space mol$

Number of mole of water = $\frac{116.92}{18}=6.5 \space mol$

Number of mole of $\ce{H2O}$ = $\frac{84.73}{100.09}=6.5 \space mol$

Note:1 mol of $\ce{CO2}$ gives 1 mol of $\ce{CaCO3}$ .

$\color{#3D99F6}{\text{4) Find coefficients b and c.}}$

$b=\frac{0.8465}{0.03527}=24$

$c=\frac{6.5}{0.03527}=184.3$

$\color{#3D99F6}{\text{5) Find x and y from balance equation in step 2.}}$

$x=b=24$

$y=\frac{2c-4x}{2}=\frac{2\times 184.3-4\times 24}{2}$

$y=136.3$

Hence $\large \frac{y}{x}=\frac{136.3}{24}=5.679$

$\huge {\color{#3D99F6}{\boxed{[\frac{y}{x}]=5}}}$

The sample of $100$ g methane hydrate yielded $116.92$ g or $\dfrac{116.92}{\color{#3D99F6}{18}} = 6.496$ mol of water (where $\color{#3D99F6}{18}$ g/mol is the molar mass of water). Therefore there were $6.496 \times 2 = 12.992$ mol of $\ce{H}$ in the sample. Therefore, $4x + 2 y = 12.993$ .

The sample also yielded $84.73$ g or $\dfrac{84.73}{\color{#3D99F6}{100.09}} = 0.847$ mol of $\ce{CaCO3}$ . Therefore, there were $0.847$ mol of $\ce{C}$ in the sample and that \(x= 0.847).

Therefore, we have:

\(\begin{cases} 4x + 2y = 12.993 & ... (1) \\ x = 10.847 & ...(2) \end{cases} \Rightarrow y = 4.802 \quad \Rightarrow \left \lfloor \dfrac{y}{x} \right \rfloor = \boxed{5}\)

Taking relative molar masses or relative atomic masses as C = 12.01, H = 1.01, O = 15.99:

16.05 x + 18.01 y = 2835.18 (1 mol) for $(CH_4)_x (H_2O)_y$ as given as first information. {Equation 1}

$[1]~(CH_4)_x (H_2O)_y$ + $[2 x]~O_2 \rightarrow$ $[x]~CO_2$ + $[2 x + y]~H_2O$

$\frac{100}{2835.18}$ mol of $(CH_4)_x (H_2O)_y$ formed $\frac{116.92}{18.01}$ mol of $H_2O$

$\implies$ 1 mol of $(CH_4)_x (H_2O)_y$ formed $\frac{116.92}{18.01} \div \frac{100}{2835.18}$ mol of $H_2O$

From chemical equation above, we obtained 2 x + y = $\frac{116.92}{18.01} \div \frac{100}{2835.18} \approx$ 184.0584373 {Equation 2}

Solving simultaneous equations above using Cramer's determinants:

x = $\frac{479.7124558}{19.97} \approx 24.02165527$ and y = $\frac{2716.222081 }{19.97} \approx 136.0151268$

So far, $\frac{y}{x} = \frac{136.0151268}{24.02165527} \approx 5.662187939$

To utilize second information,

$CO_2 + Ca(OH)_2 \rightarrow CaCO_3 + H_2O$ of coefficients of 1 on each and everyone.

$\frac{84.73}{100.09} = 0.846538116~mol$ as given to include $Ca$ for $CaCO_3$ of 100.09 g/ mol (where relative atomic mass calculated for Ca = 40.11 instead of 40.04)

0.846538116 mol of $CO_2$ is formed, which is x'. For $(CH_4)_{x'} (H_2O)_{y'}$ :

1
2
3
4
5

0.846538116 10.16692277 
0.846538116 3.420013987
4.793257909 9.682380976
4.793257909 76.64419396
~~~~~~~~~~~~99.91351169 (Approximately 100 g)

$\lfloor \frac{y}{x} \rfloor$ = 5

Answer: $\boxed{5}$