Answer:
I believe it is "Arsenenate"
<h3>
Answer:</h3>
1.93 g
<h3>
Explanation:</h3>
<u>We are given;</u>
The chemical equation;
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH = -3120 kJ
We are required to calculate the mass of ethane that would produce 100 kJ of heat.
- 2 moles of ethane burns to produce 3120 Kilo joules of heat
Number of moles that will produce 100 kJ will be;
= (2 × 100 kJ) ÷ 3120 kJ)
= 0.0641 moles
- But, molar mass of ethane is 30.07 g/mol
Therefore;
Mass of ethane = 0.0641 moles × 30.07 g/mol
= 1.927 g
= 1.93 g
Thus, the mass of ethane that would produce 100 kJ of heat is 1.93 g
Answer:
%yield of NH₃ = 30%
Explanation:
Actual yield of NH₃ = 40.8g
Theoretical yield = ?
Equation of reaction
N₂ + 3H₂ → 2NH₃
Molar mass of NH₃ = 17g/mol
Molarmass of N = 14.00
2 molecules of N = 2 * 14.00 = 28g/mol
Number of moles = mass / molar mass
Mass = number of moles * molar mass
Mass = 1 * 28.00 = 28g of N₂ (the number of moles of N₂ from the equation is 1).
From the equation of reaction,
28g of N₂ produce (2 * 17)g of NH₃
28g of N₂ = 34g of NH₃
112g of N₂ = x g of NH₃
X = (112 * 34) / 28
X = 136g of NH₃
Theoretical yield = 136g of NH₃
% yield = (actual yield / theoretical yield) * 100
% yield = (40.8 / 136) * 100
% yield = 0.3 * 100
% yield = 30%
Its chemical name is Aluminium Sulfite.
, THR CC14 formed in the first step is used as the reactant used in the second step.if 5.00 mol of CH4 reacts, what is the total amount of HCI producded. assume that C12 an HR in the presentin excess
