What mass of salt would result from the reaction between the alkali metal, and $10 \cdot m o l$ $10*mol$ $chlorine gas$ $"chlorine gas"$ ?

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What mass of salt would result from the reaction between the alkali metal, and $10 \cdot m o l$ $10*mol$ $chlorine gas$ $"chlorine gas"$ ?

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Answer 1 · 2018-02-20T18:44:32

Just under $1200 \cdot g$ $1200*g$ of salt would result....

Explanation:

We write the stoichiometric equation...

$N a (s) + \frac{1}{2} C l_{2} (g) \to N a C l (s)$ $Na(s)+1/2Cl_2(g) rarr NaCl(s)$

And so with respect to each equiv of dichlorine gas we GETS 2 equiv of salt...and since we used $10 \cdot m o l$ $10*mol$ $C l_{2}$ $Cl_2$ we get approx. $20 \cdot m o l$ $20*mol$ salt...

Thus $20 \cdot m o l \times 58.44 \cdot g \cdot m o l^{- 1} = ? ? \cdot g$ $20*molxx58.44*g*mol^-1=??*g$ ...

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What mass of salt would result from the reaction between the alkali metal, and $10 \cdot m o l$ $10*mol$ $chlorine gas$ $"chlorine gas"$ ?

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Explanation:

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What mass of salt would result from the reaction between the alkali metal, and 10*mol10⋅mol10*mol "chlorine gas"chlorine gas"chlorine gas"?

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Explanation:

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What mass of salt would result from the reaction between the alkali metal, and $10 \cdot m o l$ $10*mol$ $chlorine gas$ $"chlorine gas"$ ?