Can some one show me a simple step by step solution to solve this. The book is very confusing.
Determine the amount of NH4Cl that will be formed from the 34.0 g of NH3. Remember to use significant figures and a unit in your answer (for example, 275 g).
Print these steps.
1. Write and balance the equation.
NH3 + HCl ==> NH4Cl (The problem doesn't say it reacts with HCl and you only need to know that 1 mol NH3 is in 1 mol NH4Cl but I think an equation makes it more obvious.)
2. Convert what you have, in this case NH3, into mols. mols NH3 = grams NH3/molar mass NH3. 34.0/17 = 2.0 mols
3 Using the coefficients in the balanced equation, convert mol of what you have (NH3) in mols of what you want (in this case NH4Cl).
2.0 mols NH3 x (1 mol NH4Cl/1 mol NH3) = 2.0 mols x 1/1 = 2.0 mols NH4Cl
4. Now convert mols of what you want (NH4Cl) to g. g = mols NH4Cl x molar mass NH4Cl = 2.0 x about 53.5 = about 107.0 g NH4Cl.
Then you look at the only number given in the problem (34.0g) and that has three significant figures so the answer is allowed 3 s.f. Therefore, round that 107.0g number to 107. By the way I have estimated the numbers throughout the problem; you need to go through and do them more accurately. The molar mass NH4Cl also was estimated.
This 4-step procedure will work all simple stoichiometry problems (not limiting reagent problems---they require an extra step or two).
Don't forget to make that 107 g.
To solve this problem, you need to understand the balanced chemical equation for the reaction between NH3 (ammonia) and NH4Cl (ammonium chloride). The balanced equation is as follows:
NH3 + HCl → NH4Cl
From the equation, we can see that one mole of NH3 reacts with one mole of HCl to form one mole of NH4Cl.
To determine the amount of NH4Cl formed, we need to convert the mass of NH3 given (34.0 g) to moles using its molar mass.
The molar mass of NH3 is:
N (14.01 g/mol) + 3H (1.01 g/mol) = 17.04 g/mol
To convert grams to moles, use the conversion factor:
moles = mass / molar mass
moles of NH3 = 34.0 g / 17.04 g/mol = 1.995 moles (rounding to three significant figures)
Since the balanced equation shows a 1:1 mole ratio between NH3 and NH4Cl, we can conclude that 1.995 moles of NH4Cl will be formed.
Now, to convert the moles of NH4Cl to grams, use its molar mass:
The molar mass of NH4Cl is:
N (14.01 g/mol) + 4H (1.01 g/mol) + Cl (35.45 g/mol) = 53.49 g/mol
The mass of NH4Cl formed can be calculated using the following conversion factor:
mass = moles * molar mass
mass of NH4Cl = 1.995 moles * 53.49 g/mol = 106.88 g (rounding to three significant figures)
Therefore, the amount of NH4Cl formed from 34.0 g of NH3 is approximately 106.88 g.