4. According to the explanation given in one previous session, in this case we have the following reaction:
Zn + NO3- -> Zn(OH)42- + NH3
Our important informations to have in mind always first:
Single elements = 0 of oxidation number
Zn = 0
Oxygen = 2-
NO3- = overall charge must be -1, and we have 3 oxygens, so a -6 charge, therefore N must give a charge of +5
N (reactant) = 5+
Zn(OH)42- = overall charge -2, the charge for OH is always -1, we have 4 OH, therefore -4 of charge, to give -2 of overall, Zn must have a charge of +2
Zn (product) = 2+
OH = 1-
NH3 = 0
Now for the balancing of the reaction:
4 Zn + NO3- + 7 OH- + 6 H2O -> NH3 + 4 Zn(OH)42-
Answer:
The amount of precipitate formed would 7.175 grams of silver chloride.
Explanation:
Moles of NaCl = n
Volume of NaCl solution = 50.0 mL = 0.050 L
Molarity of the hydrogen peroxide = 2.0 M
Moles of silver nitarte = n'
Volume of silver nitrate solution = 50.0 mL = 0.050 L
Molarity of the silver nitrate = 1.0 M
According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :
of NaCl
This means that silver nitrate is in limiting amount and NaCl is in excessive amount.
So, the amount of AgCl depends upon amount of silver nitrate.
According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.
Then 0.050 moles of silver nitrate will give;
of AgCl
Mass of 0.050 moles of AgCl ;
The amount of precipitate formed would 7.175 grams of silver chloride.
Hydrogen gas and oxygen gas react to form liquid water according to the following equation:
2H₂ + O₂ → 2H₂O
a. Converting our given masses of each gas to moles, we have:
(25 g H2)/(2 × 1.008 g/mol) = 12.4 mol H2; and
(25 g O2)/(2 × 15.999 g/mol) = 0.781 mol O2.
From the equation, two moles of H2 react with every one mole of O2. To fully react with 12.4 moles of H2, as we have here, one would need 6.2 moles of O2, which is far more than what we're actually given. Thus, the oxygen is our limiting reactant, and as such it will be the first reactant to run out.
b. Since O2 is our limiting reactant, we use it for determining how much product, in this case, H2O, is produced. From the equation, there is a 1:1 molar ratio between O2 and H2O. Thus, the number of moles of H2O produced will be the same as the number of moles of O2 that react: 0.781 moles of H2O. The mass of water produced would be (0.781 mol H2O)(18.015 g/mol) ≈ 14 grams of water (the answer is given to two significant figures).
c. Since the hydrogen reacts with the oxygen in a 2:1 ratio, twice the number of moles of oxygen in hydrogen is consumed: 0.781 mol O2 × 2 = 1.562 mol H2. Since we began with 12.4 moles of H2, the remaining amount of excess H2 would be 12.4 - 1.562 = 10.838 mol H2. The mass of the excess hydrogen reactant would thus be (10.838 mol H2)(2 × 1.008 g/mol) ≈ 22 grams of hydrogen gas (the answer is given to two significant figures).
