Answer:
1. KBr
2. 2NH4OH
3. 2NO3
Explanation:
You just need to work backwords and take them step by step.
- Please let me know if these are right, if not I will correct them. Also, if you have any questions pertaining to any of this, let me know, and I will be happy to try and answer them.
Is that a question? Doesnt seem like it
The equilibrium constant (Kc) is the product of the equilibrium concentrations of the products raised to their corresponding stoichiometric coefficients divided by the reactants as well. In this case the equilibrium concentration of Cl2 which also applies to SO2 is 1.3x10^-2. The final equilibrium concentration of SO2Cl2 is 9x10^-3. Kc is then equal to 0.0188.
<h3>
Answer:</h3>
78.34 g
<h3>
Explanation:</h3>
From the question we are given;
Moles of Nitrogen gas as 2.3 moles
we are required to calculate the mass of NH₃ that may be reproduced.
<h3>Step 1: Writing the balanced equation for the reaction </h3>
The Balanced equation for the reaction is;
N₂(g) + 3H₂(g) → 2NH₃(g)
<h3>Step 2: Calculating the number of moles of NH₃</h3>
From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃
Therefore, the mole ratio of N₂ to NH₃ is 1 : 2
Thus, Moles of NH₃ = Moles of N₂ × 2
= 2.3 moles × 2
= 4.6 moles
<h3>Step 3: Calculating the mass of ammonia produced </h3>
Mass = Moles × molar mass
Molar mass of ammonia gas = 17.031 g/mol
Therefore;
Mass = 4.6 moles × 17.031 g/mol
= 78.3426 g
= 78.34 g
Thus, the mass of NH₃ produced is 78.34 g
