A sample of 79.3 g of hydrogen bromide is reacted with 80.0 g of barium hydroxide to produce barium bromide and water. Using the balanced equation below, predict which is the limiting reactant and the maximum amount in moles of barium bromide that can be produced 2HBr + Ba(OH)2 → BaBr2 + 2H2O

A. Hydrogen bromide, 0.490 Moles
B. Hydrogen bromide, 0.565 moles
C. Barium Hydroxide, 0.467 Moles
D. Barium Hydroxide, 0.624 moles

n has four pairs of electrons in the valence shell, electrons repel one another ,n has 7 protons in its nucleus, and since n has 7 protons, it must fill the second shell, giving it 4 pairs of electrons. the electrons form 3 bonds and 1 lone pair of electrons. each pair of electrons repels the other pairs, so they are equally far apart.

so all of the above

h let me think

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2