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MaRussiya [10]
3 years ago
10

If 88.0 L of natural gas, which is essentially methane (CH4), undergoes complete combustion at 720. mm Hg and 22ºC, how many gra

ms of H2O?
Chemistry
1 answer:
mr_godi [17]3 years ago
7 0

126 grams of H2O is formed.

Explanation:

Data given:

volume of the gas = 88 Liters

pressure = 720 mm Hg or 0.947 atm

temperature T = 22 Degrees or 295.15 K

R = 0.08021 atm L/mole K

n =?

The formula is used is of ideal gas law to know the number of moles of CH4 undergoing combustion.

PV = nRT

n = \frac{PV}{RT}

putting the values in the equation

   = 0.947 X 88/ 0.08021 X 295.15

n = 3.5 moles

balanced reaction for combustion of methane

CH4 + O2  ⇒ CO2 + 2H20

1 mole of CH4 undergoes combustion to form 2 moles of water

3.5 moles will give x moles of water

2/1 = x/3.5

x =  7 moles of water  (atomic mass of water = 18 gram/mole)

mass = atomic mass x number of moles

mass = 18 x 7

          =126 grams of water is formed.

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Hitman42 [59]

Explanation:

The given data is as follows.

      [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,

         V = 500 ml,   [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

   No. of moles of benzoic acid = Molarity × Volume

                         = 2 \times 0.475

                         = 0.095 mol

And, moles of NaOH present in the solution will be as follows.

    No. of moles of NaOH = Molarity × Volume

                          = 2 \times 0.025

                          = 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

         C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O

Initial:        0.095           0.05            0             0

Equlbm:  (0.095 - 0.05)  0            0.05

        pH = pK_{a} + log \frac{Base}{Acid}  

              = 4.2 + log \frac{0.05}{0.045}

              = 4.245

For,  

         HCOOH + NaOH \rightarrow HCOONa + H_{2}O

Initial:       0.2x     2(0.5 - x)               0

Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)

As,

           pH = pK_{a} + log \frac{Base}{Acid}  

          4.245 = 3.75 + log \frac{Base}{Acid}

      log \frac{Base}{Acid} = 0.5

    \frac{Base}{Acid} = 3.162

Now,

        \frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 3.162

               x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

                             = 0.036 L

                             = 36 ml               (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

                 

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