76. Diborane is a highly explosive compound formed by the reaction
NaBH4 (s) + BF3 (g) → B2H6 (g) + NaBF4 (s)
a. What mass of NaBH4, is required to form 1.00 L of diborane at STP?
b. What volume in L of BF3 is required to produce 10.0 g of NaBF4 at STP?
77. A sample of oxygen of mass 30.0 g is confined to a vessel at 0oC and 1000. torr. Then 8.00 g of N2 is pumped into a vessel at constant temperature. What will be the final pressure in the vessel (assuming only mixing with no reaction)?
79. A 6.00 L flask containing He at 6.00 atm is connected to a 3.00 L flask containing N2 at 3.00 atm and the gases are allowed to mix
a. Find the partial pressure of each gas after they are allowed to mix.
b. Find the total pressure of the mixture.
c. What is the mole fraction of Helium after the gases are mixed?
NaBH4 (s) + BF3 (g) → B2H6 (g) + NaBF4 (s)
a. What mass of NaBH4, is required to form 1.00 L of diborane at STP?
b. What volume in L of BF3 is required to produce 10.0 g of NaBF4 at STP?
77. A sample of oxygen of mass 30.0 g is confined to a vessel at 0oC and 1000. torr. Then 8.00 g of N2 is pumped into a vessel at constant temperature. What will be the final pressure in the vessel (assuming only mixing with no reaction)?
79. A 6.00 L flask containing He at 6.00 atm is connected to a 3.00 L flask containing N2 at 3.00 atm and the gases are allowed to mix
a. Find the partial pressure of each gas after they are allowed to mix.
b. Find the total pressure of the mixture.
c. What is the mole fraction of Helium after the gases are mixed?
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The molar volume of a gas at STP (assuming T = 0ºC) is 22.71 L/mol Therefore 1/22.71 moles of B2H6 is produced. 2 moles of NaBH4 produce on mole of B2H6, so the moles of NaGH4 needed is 1/(2*22.71). Multiply this times the molecular mass of NaBH4 to get the mass needed.
Divide 10g by the molecular mass of NaBH4 to get the moles; the no of moles of BF3 is the same. Multiply this time the molar volume to get the gas volume
P*V = n*R*T P/n =R*T/V = const
Find moles of O2 from the mass of O2 given, call that n1; find the moles of N2 from the mass of H2 given, add that to the moles of O2 and call that total n2.
P1/n1 = P2/n2
P2 = P1*(n2/n1), where P1 = 1000 Torr
The pressure from each gas will change according to the volume change. The helium volume changes from 6.00 L to 9.00 L so its partial pressure will be 2/3 * 6.00 = 4 atm, whild the volume of N2 changes from 3.00L to 9.00L so its partial pressure will be 1/3 * 3 = 1 atm
The total pressure is P1*V1 + P2*V2 = P3*(V1 + V2) P3 = (6.00*6.00 + 3.00*3.00)/9.00 = 5 atm. This is also the sum of the partial pressures.
The no of moles of each gas does not change after mixing. moles ~ P*V, so the mole fraction of He is 6.00*6.00/(6.00*6.00 + 3.00*3.00) = 36/45 = 4/5 = 80%
Divide 10g by the molecular mass of NaBH4 to get the moles; the no of moles of BF3 is the same. Multiply this time the molar volume to get the gas volume
P*V = n*R*T P/n =R*T/V = const
Find moles of O2 from the mass of O2 given, call that n1; find the moles of N2 from the mass of H2 given, add that to the moles of O2 and call that total n2.
P1/n1 = P2/n2
P2 = P1*(n2/n1), where P1 = 1000 Torr
The pressure from each gas will change according to the volume change. The helium volume changes from 6.00 L to 9.00 L so its partial pressure will be 2/3 * 6.00 = 4 atm, whild the volume of N2 changes from 3.00L to 9.00L so its partial pressure will be 1/3 * 3 = 1 atm
The total pressure is P1*V1 + P2*V2 = P3*(V1 + V2) P3 = (6.00*6.00 + 3.00*3.00)/9.00 = 5 atm. This is also the sum of the partial pressures.
The no of moles of each gas does not change after mixing. moles ~ P*V, so the mole fraction of He is 6.00*6.00/(6.00*6.00 + 3.00*3.00) = 36/45 = 4/5 = 80%