From the reaction of ammonia and sulfuric acid in aqueous solution
2nh3(aq)+h2so4(aq)<span>→</span> (NH4)2SO4 + H2O
will be formed.Sulfuric acid is diprotic so is able to give up 2 H+ ions.
It is an acid-base neutralisation reaction forming ammonium sulphate as the salt.
2NH3 with H2SO4 reacts in a neutralization reaction to form salt water, with ammonium sulphate left behind to crystallize after evaporation.
Balance the equation: 2Na + S --> Na2S
Using the given amount of the reactants in the reaction, calculate the amount of the product:
45.3g Na x (1 mol/22.99 g)= 1.97 mol of Na
105f S x (1 mol/ 32.06g) = 3.28 mol of S
The limiting reactant would be Na:
<span>1.97 mol Na x (1 mol Na2S/ 2 mol Na) x (78.04g/mol) = 76.87g of Na2S produced</span>
Moles of phosphoric acid would be needed : 0.833
<h3>Further explanation</h3>
Given
15 grams of water
Required
moles of phosphoric acid
Solution
Reaction(decomposition) :
H3PO4 -> H2O + HPO3
mol water (H2O :
= mass : MW
= 15 g : 18 g/mol
= 0.833
From the equation, mol ratio H3PO4 = mol H2O = 1 : 1, so mol H3PO4 = 0.833
Iodine 131 and iodine 126 are the same in the sense that, they both have the same number of electrons and protons in their atoms, it is only the number of their neutrons that is different. Iodine 131 has 78 neutrons while iodine 126 has 73 neutrons.
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