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Sodium metal (Na) reacts with water (H2O) in a single displacement reaction to produce sodium hydroxide (NaOH) and hydrogen gas (H2). When 0.8 mol of Na is placed in water. All the sodiyum metal reacts and the hydrogen gas is produced and isolated. It had been determined that 0.21 mol of H2 has been produced.

1. What is the the balance equation of the reaction between sodium metal (Na) reacts with water (H2O) producing sodium hydroxide (NaOH) and hydrogen gas (H2)?

2. What is the mole ratio/stoichiometric ratio of the reactants?

3. What is the limiting reactant?  

4. What is the excess reactant?

5. What is the theoretical yield of H2?

6. What is the actual yield of H2?

7. What is the percent yield oh H2?

Sagot :

Nicoleibay0143idn

Answer:

1) 2 Na ( s ) + 2 H2O ( l ) 2 NaOH ( aq ) + H2 ( g )

2) dividing each by the total in the whole reaction.

3) to compare the mole ratio of the amount of reactants used.

4) subtract the mass of excess reagent consumed from the total mass of excess reagent given.

5) 2 x 5 moles H = 2.5 moles of H2O

6) you multiply the percentage and theoretical yield together.

7) the experimental yield divided by theoretical yield multiplied by 100.

Explanation:

^^

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