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gtnhenbr [62]
3 years ago
15

The net ionic equation for the reaction between aqueous sulfuric acid and aqueous sodium hydroxide is ________.

Chemistry
1 answer:
Kaylis [27]3 years ago
8 0
Molecular equation:
H_{2}SO_{4}(aq) + 2NaOH(aq)--\ \textgreater \  Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Complete ionic equation:

2H^{+}+SO_{4}^{2-} + 2Na^{+} +2OH^{-}--\ \textgreater \  2Na^{+}+SO_{4}^{2-} + 2H_{2}O(l)


Near 2H^{+},SO_{4}^{2-} ,  2Na^{+} , OH^{-} 

should be written (aq).

Remove ions that do not change from both sides

2H^{+} + 2OH^{-}=2H_{2}O

Net ionic equation:

H^{+}(aq) + OH^{-}(aq)=H_{2}O(l)

Answer is D.




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In the preparation of a certain alkyl halide, 10 g of sodium bromide (NaBr), 10 mL distilled water (H20), and 9 mL 3-methyl-1-bu
Novosadov [1.4K]

Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.

The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.

Number of moles of NaBr = 10 g/103 g/mol = 0.097 moles

We can obtain the mass of 3-methyl-1-butanol from its density.

Mass = density × volume

Density of 3-methyl-1-butanol =  0.810 g/mL

Volume of  3-methyl-1-butanol = 9 mL

Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL

Mass of 3-methyl-1-butanol = 7.29 g

Number of moles of 3-methyl-1-butanol =  mass/molar mass =  7.29 g/88 g/mol = 0.083 moles

Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol

Mass of product 1-bromo-3-methylbutane = number of moles × molar mass

Molar mass of 1-bromo-3-methylbutane = 151 g/mol

Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol

= 12.53 g

Recall that % yield = actual yield/theoretical yield × 100

Actual yield of product = 6.48 g

Theoretical yield = 12.53 g

% yield = 6.48 g/12.53 g × 100

% yield = 51.7%

Learn more: brainly.com/question/5325004

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