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

1 answer:

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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Calculate the theoretical yield of alum expected from 0.9875 g of aluminum foil. assume the aluminum is the limiting reactant.

skelet666 [1.2K]

Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.

Explanation: Reaction to form alum from Aluminium is given as:

$2Al(s)+2KOH(aq.)+2H_2O(l)+4H_2SO_4(aq.)\rightarrow 2KAl(SO_4)_2(s).12H_2O(l)+3H_2(g)$

We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.

By stoichiometry,

2 moles of Al is producing 2 moles of Alum

Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol

Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol

54 g/mol of aluminium will produce 948 g/mol of alum, so

$\text{0.9875 grams of aluminium will produce}=\frac{948g/mol}{54g/mol}\times 0.9875g$