45.2 grams of aluminum nitrate are reacted with 71.9 mL of a 0.880 M solution of calcium sulfide. The balanced chemical equation can be seen below:

LaTeX: 2Al\left(NO_3\right)_3\:+\:3\:CaS\: \longrightarrow\:Al_2S_3\:+\:3Ca\le ft(NO_3\right)_22 A l ( N O 3 ) 3 + 3 C a S ⟶ A l 2 S 3 + 3 C a ( N O 3 ) 2

If 2.95 grams of precipitate is filtered out of the resulting solution, what is the actual yield?

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the answers would be

the first one and the second one

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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