How many grams of pure H2SO4 can be obtained from 250 grams of iron ore if the ore is 82% FeS2? The reactions involved are given below:4FeS2 + 11 O2 --gt; 2 Fe2O3 + 8 SO2 (95% efficient)2SO2 + O2 --gt; 2SO3 (90% efficient)SO3 + H2O --gt; H2SO4 (90% efficient)
Calculate first the mass of pure FeS2 from iron ore that will react with O2. The rest is just an inert material and will not react with O2. mass of pure FeS2 in iron ore 250 g 0f iron ore x (82/100) mass of pure FeS2 in iron ore 205 g By stoichiometric analysis: calculate the mass of SO2 produced in the first reaction since this SO2 is what we need to react with O2 in the second reaction. mass of SO2 ( mass of pure FeS2 / MW of FeS2 ) x (stoichiometric ratio of SO2 with FeS2 from the first chemical equation x MW of SO2 mass of SO2 (205 g / 119.98) x (8/4) x 64 218.70 g Take note that the mass of SO2 218.70 g is obtained at 100% efficiency. It was stated in the problem that the reaction goes 95% completion/efficiency, therefore calculate for the mass of SO2 at 95 % efficiency: mass of SO2 (95% efficiency) 218.70 x (95/100) 207.77 g Again, by stoichiometric analysis: calculate for the mass of SO3 produced from the second reaction using the mass of SO2 (95% efficiency): mass of SO3 (mass of SO2 (95% efficiency)/MW of SO2 ) x (stoichiometric ratio of SO3 and SO2 based on the second chemical reaction x MW of SO3 mass of SO3 ( 207.77 / 64 ) x (2/2) x 80 mass of SO3 259.71 g ( AT 100 % EFFICIENCY!) mass SO3 (90% efficiency) 259.71 x (90/100) mass of SO3 (90% efficiency) 233.739 g Lastly, by stoichiometric analysis again: calculate for the mass of H2SO4 produced from the third reaction using the mass of SO3 (90% efficiency): mass of H2SO4 (mass of SO3 (90% efficiency)/MW of SO3 ) x (stoichiometric ratio of H2SO4 and SO3 based on the third chemical reaction x MW of H2SO4 mass of H2SO4 ( 233.739 / 80 ) x (1/1) x 98 mass of H2SO4 286.33 g ( AT 100 % EFFICIENCY!) mass H2SO4 (90% efficiency) 286.33 x (90/100) mass of H2SO4 (90% efficiency) 257.70 g answer
Shocks is short for shock absorber. They take the big dips, potholes, rocks, etc. Springs take the every small variation in the roadway. Shocks ate typically pneumatic (air powered) or hydraulic (fluid powered). Very similar in concept to the mechanism found on screen doors that keep the doors from slamming shut. Springs are a bent pieces of metal that 'remember' their original shape and when flexed, it tries to go back to it's original shape. This allows bounce back which cancels out bounce. You can look both up. It's okay to look things up on your own.
Was interested to know the answer too
4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 (250 g) x (0.82) / (119.9762 g FeS2/mol) x (8/4) x (0.95) 3.2465 mol SO2 2 SO2 + O2 → 2 SO3 (3.2465 mol SO2) x (2/2) x (0.90) 2.9219 mol SO3 SO3 + H2O → H2SO4 (2.9219 mol SO3) x (1/1) x (0.90) x (98.0791 g H2SO4/mol) 258 g H2SO4
Calculate first the mass of pure FeS2 from iron ore that will react with O2. The rest is just an inert material and will not react with O2. mass of pure FeS2 in iron ore 250 g 0f iron ore x (82/100) mass of pure FeS2 in iron ore 205 g By stoichiometric analysis: calculate the mass of SO2 produced in the first reaction since this SO2 is what we need to react with O2 in the second reaction. mass of SO2 ( mass of pure FeS2 / MW of FeS2 ) x (stoichiometric ratio of SO2 with FeS2 from the first chemical equation x MW of SO2 mass of SO2 (205 g / 119.98) x (8/4) x 64 218.70 g Take note that the mass of SO2 218.70 g is obtained at 100% efficiency. It was stated in the problem that the reaction goes 95% completion/efficiency, therefore calculate for the mass of SO2 at 95 % efficiency: mass of SO2 (95% efficiency) 218.70 x (95/100) 207.77 g Again, by stoichiometric analysis: calculate for the mass of SO3 produced from the second reaction using the mass of SO2 (95% efficiency): mass of SO3 (mass of SO2 (95% efficiency)/MW of SO2 ) x (stoichiometric ratio of SO3 and SO2 based on the second chemical reaction x MW of SO3 mass of SO3 ( 207.77 / 64 ) x (2/2) x 80 mass of SO3 259.71 g ( AT 100 % EFFICIENCY!) mass SO3 (90% efficiency) 259.71 x (90/100) mass of SO3 (90% efficiency) 233.739 g Lastly, by stoichiometric analysis again: calculate for the mass of H2SO4 produced from the third reaction using the mass of SO3 (90% efficiency): mass of H2SO4 (mass of SO3 (90% efficiency)/MW of SO3 ) x (stoichiometric ratio of H2SO4 and SO3 based on the third chemical reaction x MW of H2SO4 mass of H2SO4 ( 233.739 / 80 ) x (1/1) x 98 mass of H2SO4 286.33 g ( AT 100 % EFFICIENCY!) mass H2SO4 (90% efficiency) 286.33 x (90/100) mass of H2SO4 (90% efficiency) 257.70 g answer
If you call the 'Shock absorber' by its proper name which is a Damper it is much clearer. A spring allows the suspension to move up and down and smooth the ride. The trouble with springs is they carry on oscillating. A dumper literally damps the oscillation. Disconnect the dampers on the back of a car. Then bounce the suspension. Reconnect the dampers and see the difference.
Shocks is short for shock absorber. They take the big dips, potholes, rocks, etc. Springs take the every small variation in the roadway. Shocks ate typically pneumatic (air powered) or hydraulic (fluid powered). Very similar in concept to the mechanism found on screen doors that keep the doors from slamming shut. Springs are a bent pieces of metal that 'remember' their original shape and when flexed, it tries to go back to it's original shape. This allows bounce back which cancels out bounce. You can look both up. It's okay to look things up on your own.
Was interested to know the answer too
4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 (250 g) x (0.82) / (119.9762 g FeS2/mol) x (8/4) x (0.95) 3.2465 mol SO2 2 SO2 + O2 → 2 SO3 (3.2465 mol SO2) x (2/2) x (0.90) 2.9219 mol SO3 SO3 + H2O → H2SO4 (2.9219 mol SO3) x (1/1) x (0.90) x (98.0791 g H2SO4/mol) 258 g H2SO4
If you call the 'Shock absorber' by its proper name which is a Damper it is much clearer. A spring allows the suspension to move up and down and smooth the ride. The trouble with springs is they carry on oscillating. A dumper literally damps the oscillation. Disconnect the dampers on the back of a car. Then bounce the suspension. Reconnect the dampers and see the difference.
