In: Chemistry
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
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 %