Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g
Answer:
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but PhC2H5 + O2 = PhOH + CO2 + H2O will; Compound states [like (s) (aq) or (g)] are not required. If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.
Explanation:
Answer:
3.81 g Pb
Explanation:
When a lead acid car battery is recharged, the following half-reactions take place:
Cathode: PbSO₄(s) + H⁺ (aq) + 2e⁻ → Pb(s) + HSO₄⁻(aq)
Anode: PbSO₄(s) + 2 H₂O(l) → PbO₂(s) + HSO₄⁻(aq) + 3H⁺ (aq) + 2e⁻
We can establish the following relations:
- 1 A = 1 c/s
- 1 mole of Pb(s) is deposited when 2 moles of e⁻ circulate.
- The molar mass of Pb is 207.2 g/mol
- 1 mol of e⁻ has a charge of 96468 c (Faraday's constant)
Suppose a current of 96.0A is fed into a car battery for 37.0 seconds. The mass of lead deposited is:
The SI base units and their physical quantities are the metre for measurement of length, the kilogram for mass, the second for time, the ampere for electric current, the kelvin for temperature, the candela for luminous intensity, and the mole for amount of substance.
