<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
The elements in group 7, which fluorine is in also, would have similar properties due to the valence electrons and the configuration. <span />
The maximum mass of B₄C that can be formed from 2.00 moles of boron (III) oxide is 55.25 grams.
<h3>What is the stoichiometry?</h3>
Stoichiometry of the reaction gives idea about the relative amount of moles of reactants and products present in the given chemical reaction.
Given chemical reaction is:
2B₂O₃ + 7C → B₄C + 6CO
From the stoichiometry of the reaction, it is clear that:
2 moles of B₂O₃ = produces 1 mole of B₄C
Now mass of B₄C will be calculated by using the below equation:
W = (n)(M), where
- n = moles = 1 mole
- M = molar mass = 55.25 g/mole
W = (1)(55.25) = 55.25 g
Hence required mass of B₄C is 55.25 grams.
To know more about stoichiometry, visit the below link:
brainly.com/question/25829169
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Answer:
It's 10 cenimeters per second
