For the reaction 3NaBH4(s) + 4BF3(g) --> 2B2H6(g) + 3NaBF4(s) , how many grams of NaBH4 are required to form 1.75L of B2H6 @ 25 degrees C, 1 atm?
First let's use the ideal gas law (PV=nRT) to find out how many moles of B2H6 are contained in 1.75 liters of the gas. The temperature in Kelvin is 298 K, the pressure is 1 atm, and the constant is 0.0821 L atm/mol K.
n=PV/RT=[(1 atm)(1.75 L)]/[(0.0821 L atm/mol K)(298 K)=0.071 moles B2H6.
From the chemical equation we see that 3 moles of sodium borohydride (NaBH4) are required for every 2 moles of B2H6. We can use this to calculate the number of moles of NaBH4 needed:
0.071 moles B2H6*(3 moles NaBH4/2 moles B2H6)=0.107 moles NaBH4.
We can now use the molecular weight of NaBH4 (37.83 g/mol) to convert the moles to grams:
0.107 moles NaBH4*(37.83 g/mol)=4.06 g NaBH4.
We see that 4.06 grams of NaBH4 are needed to make 1.75 liters of B2H6.