Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:
Best regards.
T = 14400 s
26.5 x 14400=381600 C
381600/96500=3.95 Faradays
Cu2+ + 2e- = Cu
3.95 faradays ( 1 mol/ 2 Faradays) = 1.97
mass = 1.97 x 63.55 g/mol=125 g
moles Au = 33.1 / 196.967 g/mol=0.168
Au+ + 1e- = Au
0.168 ( 1 Faraday/ 1mol)= 0.168 Faraday
0.168 x 96500=16217 Coulombs
16217 / 5.00=3243 s => 54 min
Answer:
Mass of C₂H₄N₂ produced = 3.64 g
Explanation:
The balanced chemical equation for the reaction is given below:
3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)
From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O
Molar masses of the compounds are given below below:
CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol
Comparing the mole ratios of the reacting masses;
CH₄ = 1.65/16 = 0.103
CO₂ = 13.5/44 = 0.307
NH₃ = 2.21/17 = 0.130
converting to whole number ratios by dividing with the smallest ratio
CH₄ = 0.103/0.103 = 1
CO₂ = 0.307/0.103 = 3
NH₃ = 0.130/0.103 = 1.3
Multiplying through with 5
CH₄ = 1 × 5 = 5
CO₂ = 3 × 5 = 15
NH₃ = 1.3 × 5 = 6.5
Therefore, the limiting reactant is NH₃
8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂
Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂
Mass of C₂H₄N₂ produced = 3.64 g
