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?

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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.

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