Can someone give me some directions as to how to reach the answer for this problem:
A 0.608g sample of fertilizer contained nitrogen as ammonium sulfate,
(NH4)2 SO4(s)+2NaOH(aq)-->Na2SO4(aq)+ 2H2O(l)+2NH3(g)
The ammonia was collected in 46.3 mL of 0.213 M HCL (hydrochloric acid), with which it reacted.
NH3(g)+HCL(aq)-->NH4Cl(aq)
This solution was titrated for excess hydrochloric acid with 44.3 mL of 0.128 M NaOH
NaOH(aq)+HCL(aq)NaCL(aq)+H2O(l)
What is the percentage of nitrogen in the fertilizer?
The answer is 9.66%
1. Find mol of NaOH:
NaOH: 44.3mL * 0.128 mol / 1 L -> 5,6704 * 10⁻³ mol NaOH
Since proportion NaOH and HCl is 1:1, 5,67 * 10⁻³ mol HCl left, after previous NH₃ + HCl reaction.
2. Find initial mol of HCl
HCl: 46.3 mL * 0.213 mol / 1 L -> 9,86 * 10⁻³ mol HCl
3. Find reacted HCl mol: 9,86 * 10⁻³ - 5,67 * 10⁻³ = 4,19 * 10⁻³ mol HCl
NH₃:HCl = 1:1, thus 4,19 * 10⁻³ mol NH₃ was in the first reaction
4. Find mass of nitrogen in 4,19 * 10⁻³ mol NH₃ (4,19 * 10⁻³ mol N)
Molar mass N = 14 g/mol
4,19 * 10⁻³ mol N * 14 g N / 1 mol N -> 58.66 * 10⁻³ g N
5. Find mass percentage
58.7mg / 608mg = 9.65%
To find the percentage of nitrogen in the fertilizer, we need to calculate the number of moles of nitrogen in the sample and then divide it by the total mass of the sample. Here are the steps to reach the answer:
1. Calculate the number of moles of NH3 produced:
From the balanced equation (NH4)2SO4(s) + 2NaOH(aq) --> Na2SO4(aq) + 2H2O(l) + 2NH3(g), we can see that for every 1 mole of (NH4)2SO4, 2 moles of NH3 are produced.
Therefore, the moles of NH3 produced can be calculated using the molarity (M) and volume (V) of HCl used in the reaction:
moles of NH3 = Molarity x Volume
= 0.213 M x 0.0463 L (converted mL to L)
= 0.0098579 moles
2. Calculate the number of moles of HCl neutralized:
Since 2 moles of NH3 react with 2 moles of HCl, the moles of HCl neutralized by NH3 can be calculated as follows:
moles of HCl neutralized = 0.0098579 moles
3. Calculate the number of moles of excess HCl:
Since the volume of NaOH used in the titration is given, we can calculate the moles of NaOH used:
moles of NaOH = Molarity x Volume
= 0.128 M x 0.0443 L (converted mL to L)
= 0.0056704 moles
From the balanced equation NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(l), we can see that 1 mole of NaOH reacts with 1 mole of HCl. Therefore, there are also 0.0056704 moles of HCl neutralized.
To find the moles of excess HCl, subtract the moles of HCl neutralized by NH3 from the total moles of HCl:
moles of excess HCl = moles of HCl used - moles of HCl neutralized
= 0.0056704 moles - 0.0098579 moles
= -0.0041875 moles (negative sign indicates excess HCl)
4. Calculate the moles of nitrogen in the fertilizer:
Since 2 moles of NH3 are produced from 1 mole of (NH4)2SO4, we can calculate the moles of nitrogen obtained from NH3 as follows:
moles of nitrogen = 2 x moles of NH3
= 2 x 0.0098579 moles (moles of NH3 obtained in step 1)
= 0.01972 moles
5. Calculate the mass of nitrogen:
The molar mass of nitrogen (N) is approximately 14.01 g/mol.
mass of nitrogen = moles of nitrogen x molar mass of nitrogen
= 0.01972 moles x 14.01 g/mol
= 0.275992 g
6. Calculate the percentage of nitrogen in the fertilizer:
The mass of the sample is given as 0.608 g.
percentage of nitrogen = (mass of nitrogen / mass of sample) x 100
= (0.275992 g / 0.608 g) x 100
= 45.39%
Therefore, the percentage of nitrogen in the fertilizer is 45.39%.
Sorry, but your wording is quite confusing. How would you convert from NH3 to (NH4)2SO4 then to moles N (not N2), I don't know if I understand your reasoning...since you already have the sample mass and molar mass of (NH4)2SO4 so therefore you can work out how many moles it has- so how do the two relate?
How much HCl was used to collect the NH3? That will be moles HCl = M x L.
How much excess HCl was there. That will be the excess titration with NaOH. Moles NaOH = moles HCl excess = L NaOH x M NaOH.
Subtract the amount HCl in which the NH3 was collected from the total HCl to give the moles NH3 from the sample.
Then convert moles NH3 to moles (NH4)2SO4 then to moles N (not N2) using the coefficients in the balanced equation. From there go to grams N, then percent N from
percent N = [(grams N)/grams sample]*100
Post your work if you get stuck.