I'm going to suppose you want the adjusted chemical reaction, using the formulas of the compounds. You can see it in the image attached.
1) Chemical reaction (thermal decomposition)
2NaHCO3 (s) ---> Na2CO3 (s) + H2O (g) + CO2(g)
2) Reasoning
The lost of mass is due to the lost of the gases H2O and CO2.
So, you can calculate the mass of Na2CO3 obtained from 1.000 g NaHCO3, and the difference will be the mass lost.
2) Convert 1.000 g of NaHCO3 to number of moles
molar mass NaHCO3: 1*23g/mol + 1*1g/mol + 1*12g/mol + 3*16g/mol = 84 g/mol
number of moles = mass in grams / molar mass = 1.000 g / 84 g/mol = 0.01190 moles
3) Use therotecial molar ratios:
2 moles NaHCO3 : 1 mol Na2CO3
=> 0.01190 mol NaHCO3 / x = 2 mol NaHCO3 / 1mol Na2CO3
=> x = 1mol Na2CO3 * 0.01190 mol NaHCO3 / 2 mol NaHCO3
=> x = 0.00595 mol Na2CO3
4) Convert 0.0595 mol Na2CO3 to mass
molar mass Na2CO3: 2*23g/mol + 1*12g/mol + 3*16g/mol = 106 g/mol
mass in grams = number of moles * molar mass = 0.00595 mol * 106 g/mol = 0.6307 g
5) lost mass
1.000g - 0.6307g = 0.3693 g
Answer: 0.3693 g
<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
