according to the equation 2na + 2h₂o → 2naoh+h₂, what mass of na is required to yield 22.4 l of h₂ at stp…

Explanation:

Step1: Determine moles of H₂ at STP

At STP, 1 mole of any gas occupies 22.4 L. So, for 22.4 L of H₂, the number of moles of H₂, $n_{H_2}=\frac{22.4\ L}{22.4\ L/mol}= 1\ mol$.

Step2: Use mole - ratio from the balanced equation

The balanced equation is $2Na + 2H_2O\rightarrow2NaOH + H_2$. The mole - ratio of Na to H₂ is 2:1. So, if $n_{H_2}=1\ mol$, then the number of moles of Na, $n_{Na}=2\times n_{H_2}=2\ mol$.

Step3: Calculate mass of Na

The molar mass of Na is 22.99 g/mol. Using the formula $m = n\times M$, where $n$ is the number of moles and $M$ is the molar mass. So, $m_{Na}=n_{Na}\times M_{Na}=2\ mol\times22.99\ g/mol\approx46.0\ g$.

Answer:

46.0 g