When 23 g of Na reacts with 1 mole of propy | vedclass

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(B) The reaction between propyne and sodium is given by: $CH_3-C\equiv CH + Na \longrightarrow CH_3-C\equiv C^{-}Na^{+} + 0.5 \ H_2$ Given that $23 \

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$0.5$

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(B) The reaction between propyne and sodium is given by: $CH_3-C\equiv CH + Na \longrightarrow CH_3-C\equiv C^{-}Na^{+} + 0.5 \ H_2$ Given that $23 \ g$ of $Na$ is equal to $1 \ mole$ $(n = \frac{23 \ g}{23 \ g/mol} = 1 \ mole)$. According to the stoichiometry of the reaction,$1 \ mole$ of $Na$ reacts with $1 \ mole$ of propyne to produce $0.5 \ mole$ of $H_2$ gas.

When $23 \ g$ of $Na$ reacts with $1 \ mole$ of propyne,how many moles of $H_2$ gas will be released?

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