balance the following equation
NaBh4 (s) + BF3 (g) - B2H6 (g) + NaBF4 (s)
The given chemical equation is
NaBH4 (s) + BF3 (g)
B2H6 (g) + NaBF4 (s)
We can balance the given equation by adjusting stoichiometric coefficients of the chemical species to obtain the same number of atoms of elements on both sides of the equation.
Then adjusting the coefficients,
3 NaBH4 (s) + 4 BF3 (g) 2
B2H6 (g) + 3 NaBF4 (s)
Now checking the number of atoms of elements on both sides. 3 sodium atoms on both sides of the equation, 7 boron atoms each on reactant and product side, 12 hydrogen atoms and 12 fluorine atoms on both sides of the equation. So the equation got balanced.
The balanced equation is 3 NaBH4 (s) + 4
BF3 (g) 2
B2H6 (g) + 3 NaBF4 (s)
balance the following equation NaBh4 (s) + BF3 (g) - B2H6 (g) + NaBF4 (s)
Balance the following equation. B2H6(g) + O2(g) -B203(s) + H2O(g) What is the stoichiometric coefficient for oxygen? 04 Previous Page Next Page Page 5
2 NaBH4(aq) + H2SO4(aq) → 2 H2(g) + Na2SO4(aq) + B2H6(g) What volume, in mL, of a 0.501 M solution of NaBH4 is required to produce 0.497 g of B2H6? H2SO4 is present in excess.
Consider the reaction: 2 NaBH4(aq) + H2SO4(aq) → 2 H2(g) + Na2SO4(aq) + B2H6(g) What volume, in mL, of a 0.540 M solution of NaBH4 is required to produce 0.525 g of B2H6? H2SO4 is present in excess.
Consider the reaction: 2 NaBH4(aq) + H2SO4(aq) → 2 H2(g) + Na2SO4(aq) + B2H6(g) What volume, in mL, of a 0.529 M solution of NaBH4 is required to produce 0.547 g of B2H6? H2SO4 is present in excess. mL
Consider the reaction: 2 NaBH4(aq) + H2SO4(aq) → 2 H2(g) + Na2SO4(aq) + B2H6(g) What volume, in mL, of a 0.515 M solution of NaBH4 is required to produce 0.541 g of B2H6? H2SO4 is present in excess. mL
Fill in the blanks to balance the following equation: B(s) + F2 (g) → BF3 (8) Hint Remember the number of each type of atom must be the same on both sides of the equation. Answered 2 attempts left Resubmit Practice Problem 5.101 Homework – Unanswered Fill in the Blanks Fill in the blanks to balance the following equation: NaClO3(s) - - Naci (3) + NaCl(s) 0 0.0
Diborane, B2H6, can be produced by the following reaction: 〖2NaBH〗_4 (s)+H_2 〖SO〗_4 (aq)→〖2H〗_2 (g)+〖Na〗_2 〖SO〗_4 (aq)+B_2 H_6 (g) Calculate the volume, in milliliters, of 0.0875-M H2SO4 needed to completely react with 1.35 g NaBH4.
Consider the following thermal equations: 2B (s) + 3H2(g) ⟶ B2H6 (g) ΔH = +36kJ/mol 2B (s) + 3/2O2 (g) ⟶ B2O3 (s) ΔH = −1273 kJ/mol H2 (g) + 1/2O2 (g) ⟶ H2O (l) ΔH = −286 kJ/mol H2O (l) ⟶ H2O (g) ΔH = +44 kJ/mol Calculate ΔH for the combustion of borane, B2H6 (g) + 3O2 (g) ⟶ B2O3 (s) + 3H2O (g)
Using Hess's Law, calculate ΔH°R Equation: B2H6 (g) + 6 Cl2 (g) --> 2BCl3 (g) + 6 HCl (g) Given these 3 equations: (please show all work) BCl3 (g) + 3H2O (l) --> H3BO3 (g) + 3HCl (g) (ΔH°R = -112.5 KJ/mol BCl3) B2H6 (g) + 6H2O (l) --> 2H3BO3 (g) + 6H2 (g) (ΔH°R = -493.4 KJ/mol B2H6) H2 (g) + Cl2 (g) --> 2HCl (g) (ΔH°R = -184.6 KJ/mol H2)
1)Consider the reaction B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g) ∆H = -2035 kJ/mol Calculate the amount of heat released when 37.1 g of diborane is burned. 2) Consider the reaction B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g) ∆H = -2035 kJ How much heat is released when a mixture of 9.71 g B2H6 and 1.53 g O2 is burned? 3)Consider the reaction B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g) ∆H = -2035 kJ...