Which, if any, of the following equations is or are correctly balanced?
A . 4LiH + AlCl3 → 2LiAlH4 + 2LiCl
B. Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O
C. 4BF3 + 3H2O → H3BO3 + 3HBF4
D. 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O
Any help or explination is greatly appreciated!!!
2006-11-20 12:31:05 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
A . 4LiH + AlCl3 → 2LiAlH4 + 2LiCl <-- there's only 1 alluminum so it can't yield 4 in the product
B. Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O <-- there are 1 sn, 4 H, 2 S and 8 O ( multiply the coefficient [2] by the subscript [4])
it matches in the product
CORRECT
C. 4BF3 + 3H2O → H3BO3 + 3HBF4< same goes here, just write out how many of each element are in the beginning of the equation and in the product and compare
CORRECT
D. 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O <-- this one is hard to read, I'm assuming that I is not an L because it's followed by a subscript, so it would be iodine
but either way, this one is wrong because there are 5 C/Cl in the product which is wrong
2006-11-20 12:53:52 · answer #1 · answered by Leger de Main 2 · 0⤊ 0⤋
B, C and D are correctly balanced. A is not.
To balance a chemical reaction, just count the number of every atom in the reactant side and make sure it is equal to the number of atoms in the product side.
For example in C. Four (4) Boron atoms in reactants equal 1 and 3 Boron in the product side. 4x3 Flourine atoms in the reactant are equal to 3x4 Flourine in the product side.
Similarly, there are 6 H and 3 O in the reactant side equal 3+3 = 6 H and 3 O in product side.
In D. it is Chlorine is in the form of Chlorine gas, which is diatomic. It is not Chlorine and Iodine.
2006-11-20 12:56:45 · answer #2 · answered by Aldo 5 · 0⤊ 0⤋
D) Its called stoichiometry
Tally of the number of coefficients on each side for each reactant and product. Then make sure that the numbers and kinds of atoms are same on both sides of the equation.
2006-11-21 09:53:30 · answer #3 · answered by yessenia 3 · 0⤊ 0⤋
D. each atom in the reactant has to be present in the product.
2006-11-20 12:37:03 · answer #4 · answered by Dan 2 · 0⤊ 0⤋
a million) mulitiply each and every term contained in the 2nd equation by 4; subtract the ensuing equation from the 1st one. 2) multiply each and every term contained in the 1st equation by 6/5; upload the ensuing equation to the 2nd. 3) x = 0.01724 4) y = 0.2931
2016-10-04 04:45:08 · answer #5 · answered by lyon 4 · 0⤊ 0⤋
D, each element should be equally represented in the reaction and the product.
2006-11-20 12:42:20 · answer #6 · answered by kmccrae8403 2 · 0⤊ 0⤋