According to the following reaction, how many grams of ammonium chloride are needed to form 30.8 grams of hydrochloric acid? ammonium chloride (aq) ammonia (g) + hydrochloric acid (aq)

=27.876g of Calcium Chloride

6.023×1023 molecules of ammonia = 1 mole ammonia7.81×1024 molecules of ammonia = (1×7.81×1024) / 6.023×1023                                                        = 12.9612.96 moles of ammonia was produced.

NH3 = 17 g/molepH = 9–––> [H+] = 10^-9 M ––––> [OH–] = 10^-5 MKa*Kb = Kwater(5.6×10^-10)*Kb = 10^-14Kb = 1.8×10^-5 ( Kb of NH3)[OH–] = nNH3*Kb / nNH4Cl10^-5 =...

It's an acid - base reaction because there is no change in oxidation states of the elements.CO2 - acidic, produces acid with water, NH3 - basic, proces base with water....

The chemical equation for thi reaction is Al(OH)3 + 3HCl = AlCl3 + 3H2O MW (HCl) = MW (H) + MW (Cl) = 1.0 + 35.5 = 36.5 g/mol MW...

Solution:The molar mass of NH3 is 17.031 g/mol.Hence,a) Mole to Grams conversion:2.4 mol NH3 × (17.031 g NH3 / 1 mol NH3) = 40.8744 g NH3 = 40.87 g NH3Answer...

6.023×1023 molecules of ammonia = 1 mole ammonia7.81×1024 molecules of ammonia = (1×7.81×1024) / 6.023×1023                                                        = 12.96 mole ammonia (Answer)12.96 mole of ammonia was produced.

Given,Number of molicule of amomonia produce = 7.19 x 1024 molecules We know, 6.023 x 1023 molicule = 1 mole 1 molicule = (1/(6.023x1023)) mole 7.19 x 1024 molecules...

82.29 litres

355g