Calculate the mass of water produced when 3.8 x 103 grams of C6H12O6 are reacted

Question

Calculate the mass of water produced when 3.8 x 103 grams of C6H12O6 are reacted with excess oxygen gas (O2). Assume the products are those expected for a typical hydrocarbon combustion, CO2 and H2O.

The empirical formula for a compound that is 36.80% sodium and the balance (the rest) selenium is NaxSey where the ratio x:y is

10.00 grams of C2H6 reacts with 10.00 grams of elemental oxygen in a typical combusion reaction. Write down the balanced reaction and determine which reactant is the limiting reagent.

In the reaction of aqueous solutions of strontium sulfide with magnesium sulfate the correct complete ionic equation is

please hepl i dont understand

Explanation / Answer

1) C6H12O6 + 6O2   -------> 6 CO2 + 6H2O

as 1 mole of   C6H12O6 react with excess of O2 . the limiting reagent is   C6H12O6

1 mole of C6H12O6 reacts to produces 6 mole of water

180 g of C6H12O6 reacts to produces = 6 *18 g of water

1 g of C6H12O6 reacts to produces = 108 / 180 g of water

3800 g C6H12O6 reacts to produces = 108 / 180* 3800 =    2280 g of water

2) % of element atomic mass moles of element simplest molar ratio

sodium 36.80 23 36.80/   23 =1.6     1.6 /0.8 = 2

selenium 63.2 78.96   63.2 / 78.96 = 0.8 0.8 /0.8 = 1

NaxSey = Na2Se1 =

the ration of x:y = 2:1

3) 2 C2H6 + 7 O2 -------> 4CO2 + 6H20

   as 2 mole of C2H6 combines with 7 moles of O2

2* 30 g of C2H6 combines with 7 * 32 g of  O2

10 g of C2H6 requires  7 * 32 / 2* 30 * 10 = 37.4 g of O2

as O2 is in excess therefore C2H6 is he limiting reagent

4) :  SrS (aq)  + MgSO4 (aq) ----> SrSO4 (s)   + MgS (aq)

total ionic = Sr 2+ (aq) + S2- (aq) + Mg2+(aq)+ SO42- (aq) ----->   SrSO4 (s)+   Mg2+(aq)+   S2- (aq)

net ionic = Sr 2+ (aq) +  SO42- (aq) ---->   SrSO4 (s)

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