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SVETLANKA909090 [29]

3 years ago

10

The balanced chemical equation for the combustion of butane is: 2C2H2 + 5O2 CO2 + 2H2O 2CH4 + 5O2 2CO2 + 4H2O 2C4H10 + 13O2 8CO2

+ 10H2O C4H10 + 12O2 4CO2 + 5H2O

1 answer:

SCORPION-xisa [38]3 years ago

3 0

Butane is C₄H₁₀.

$C_4H_{10} + O_2 \to CO_2 + H_2O \\ \\
\hbox{balance carbon and hydrogen on the right-hand side:} \\
C_4 H_{10} + O_2 \to 4 \ CO_2 + 5 \ H_2O \\ \\
\hbox{balance oxygen on the left-hand side:} \\
C_4 H_{10} + \frac{13}{2} \ O_2 \to 4 \ CO_2 + 5 \ H_2O \\ \\
\hbox{multiply by 2 to get rid of the fraction:} \\
2 \ C_4H_{10} + 13 \ O_2 \to 8 \ CO_2 + 10 \ H_2O$

The balanced equation is 2 C₄H₁₀ + 13 O₂ <span>→</span> 8 CO₂ + 10 H₂O.

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Explanation:

Depression in freezing point:

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where,

$T_f$ = freezing point of solution = ?

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$k_f$ = freezing point constant of water = $1.86^0C/m$

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m = molality

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$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

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$T_b$ = boiling point of solution = ?

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$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

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