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1) How many grams of CO are needed to react with an excess of Fe2O3 to produce 209.7g Fe? Show your work.
Fe2O3 (s) + 3CO (g) ->3CO2 (g) + 2Fe (s)

2) Solid sodium reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of hydrogen gas are formed when 48.7 g of sodium are added to water?
2Na + 2H2O -> 2NaOH + H2

3) Consider the following reaction:
2H2S (g) + 3O2 (g) ->2SO2 (g) + 2H2O (g)
If O2 was the excess reagent, 8.3 mol of H2S were consumed, and 137.1 g of water were collected after the reaction has gone to completion, what is the percent yield of the reaction?


Please show all work thank you.

Respuesta :

SASSAS
1.) is157.7 g
moles Fe = 209.7 g/ 55.847 g/mol=3.75 
the ratio between Fe and CO is 2 : 3 
moles CO = 3.75 x 3 /2 =5.63 
mass CO = 5.63 mol x 28.01 g/mol=157.7 g

2.) is 1.06 moles 
48.7/23 = 2.12 moles sodium 
2.12/2 x 24 = 25.44dm^3 hydrogen = 1.06 moles 
1.06 X 6.02x10^23 = 1.204x10^24 molecules of hydrogen.

3.) is 91.8 
8.3 moles H2S x (2 moles H2O / 2 moles H2S) = 8.3 moles H2O = theoretical amount produced. 8.3 moles H2O x (18.0 g H2O / 1 mole H2O) = 149 g H2O produced theoretically. % yield = (actual amount produced / theoretical amount) x 100 = (137.1 g / 149 g) x 100 = 91.8 
Bladerunner1976
mol Fe = 209.7/56 = 3.744 mol Fe
mol CO = 3.744 * 3/2 (mol ratio) = 5.616 mol CO
molecular weight of CO = 28
5.616 * 28 = 157.28 g CO needed

48.7/23 = 2.11 mol Na
2.11 * 1/2 = 1.055 mol H2
1.055 * 6.023*10^23 = 6.351 x 10^23 molecules of H2

8.3 * 2/2 = 8.3 mol H2O that could be produced
137.1/18 = 7.166 mol actually produced
7.616/8.3 * 100% = 91.75% produced (% yield)

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