Hydrogen gas is used as a fuel instead of fossil fuels. Storage of hydrogen is a

Question

Hydrogen gas is used as a fuel instead of fossil fuels. Storage of hydrogen is a challenge. One avenue of research involves chemical storage such as incorporating hydrogen as part of a metal hydride. In the following reaction sodium borohydride, NaBH4, is used as a means for storing and later generating the hydrogen gas in situ where it is needed as a fuel. Application of heat and use of a catalyst drives the reaction forward. NaBH4(s) + 2 H2O(l) NaBO2(s) + 4 H2(g) To test the efficiency of the reaction, 0.11 g of sodium borohydride is catalytically decomposed and 244 mL of hydrogen gas is collected over water at 30.0 °C under an atmospheric pressure of 752 mmHg. Calculate the % yield of the reaction

Explanation / Answer

Calculation of number of moles of H2 gas produced:-

We know that PV = nRT

Where

P = Pressure = 752 mm Hg = (752 / 760) atm                Since 1 atm = 760 mm Hg

   = 0.989 atm

V = volume = 244 mL = 0.244 L

n = number of moles = ?

R = gas constant = 0.0821 Latm / (mol-K)

T = temperature = 30.0 oC = 30.0+273 = 303.0 K

Plug the values we get

n = (PV) / (RT)

   = (0.989 x 0.244) / ( 0.0821x303)

   = 0.0097 moles

So mass of H2 produced ,m = number of moles x molar mass

                                          = 0.0097 mol x 2(g/mol)

                                          = 0.0194 g -----------------> this is the actual yield

Given mass of NaBH4(sodium borohydrate) is = 0.11 g

Molar mass of NaBH4 = 23+10.8 +(4x1) = 37.8 g/mol

So number of moles of NaBH4 , n = mass/molarmass

                                                 = 0.11 g / 37.8 (g/mol)

                                                = 0.0029 mol

                           NaBH4(s) + 2 H2O(l) NaBO2(s) + 4 H2(g)

According to the balanced equation,

1 mole of NaBH4 produces 4 moles of H2

0.0029 mole of NaBH4 produces M moles of H2

M = (0.0029 x 4) / 1

   = 0.0116 moles of H2

So the mass of H2 produced , m ' = number of moles x molar mass

                                                 = 0.0116 mol x 2 (g/mol)

                                                  = 0.0232 g    -----------------> this is the theoretical yield

So percent yield = ( actual yield /theoretical yield) x100

                        = ( 0.0194 / 0.0232) x 100

                        = 83.6 %

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