sample of iron (Fe) was burned in a crucible at high heat with powder sulfur(S)t
ID: 695206 • Letter: S
Question
sample of iron (Fe) was burned in a crucible at high heat with powder sulfur(S)t. After cooling iron sulfide formed was weighed. Calculated results of the mass of iron( Fe given below. ) and mass of sulfur (s) Mass of Fe = 0.5373 g Mass of S-04627 g and Calculate Empirical formula of iorn sulfide formed in this experiment. Remember iron Show all steps of calculations. (4 points) forms two sulfides. 4 A sample of 1.35 g of an impure sample of KHP was titrated with 0.110 M Na0H standardized solution The volume of NaOH needed to neutralize was 36.00 ml of NaOH. Calculate the percent of KHP in this sample (Molar mass of KHP is 204.2g) Show all work with units and c.v.and f.v. (4 points) What is the volume of 0.250 M NaOH needed to neutralize 15.20 mL of 0.125 M HCI? Show all work. (3 points) NaOH (aq) HCI (aq)NaCI (ag) HO I) Equation and rearranged EQ Calculation Steps an experiment to calculate heat of neutralization reaction using a calorimeter using a coffee cup, a of students added 1.969 g of solid NaOH to 100.0 ml (100.0 g) of 0.50 M HCI splution. The initial erature of the solution was 22.5 °C and final temperature measured was 33.9. Calculate the hea alization (a) in this reaction. (use SH of water as 4.181/8). Give final value of q in Finish with formula of q below(1 pt.) Calculation step with units (2 pts)
Explanation / Answer
a)
mol of Fe = mass/MW = 0.5373/55.8 = 0.009629
mol of S = masS/MW = 0.4627/32 = 0.01445
ratio = 0.01445/0.009629 = 1.5
1.5 S : 1Fe
FeS1.5 = FeS3
Q4
mol o fKHP = mass/MW = 1.35/204.2 = 0.006611
mol of NAOH = MV = 0.11*36*10^-3 = 3.96*10^-3
mol ofKHP = 3.96*10^-3
% KHP = mass of KHP / Total mass * 100% = (3.96*10^-3)/0.006611*100 = 59.900%
Q3
mmol of HCL = MV = 0.125*15.20 = 1.9
mmol of NAOH = MV = 0.25*V
V = 1.9/0.25 = 7.6 mL
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.