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aivan3 [116]
3 years ago
9

Write the balanced equation for the reaction that occurs when methanol, CH3OH (I) is burned in air. What is the coefficient of m

ethanol in the balanced equation? (A) 1 (B) 3 (C) 2 (D) 4 (E) 3/2
Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

(C) 2

Explanation:

Methanol is an organic compound which burns with oxygen to produce carbon dioxide and water.

2CH_3OH_{(l)}+3O_2_{(g)}\rightarrow 2CO_2_{(g)}+4H_2O_{(g)}

Multiply coefficients times subscripts for each element in the formula, We get that there are 2 number of atoms of the carbon, 8 number of atoms of the hydrogen, and 8 number of atoms of the oxygen on both sides of equation. Hence, it is balanced.

<u>The coefficient of methanol from the balanced reaction is 2.</u>

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4 years ago
How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?
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Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

H2SO3\leftrightarrow H^{+}+HSO3^{-} --------(1)

where,  Ka1 = 1.5 x 10–2  or pKa1 = 1.824

HSO3^{-}\leftrightarrow H^{+}+SO3^{2-} --------(2)

where,  Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

i.e. \ 5.13\  moles/L*0.385L = moles\  NaOH\\therefore, \ moles\  NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L

Step 2:

For the second equivalence point setup an ICE table:

                  HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}

Initial           1.98                    ?                                       0

Change      -x                       -x                                       x

Equil           1.98-x                 ?-x                                    x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles

Volume of NaOH required is:

\frac{0.634\ moles}{0.615 moles/L}=0.389L

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

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