# 2"Na" + 2"H"_2"O"->2"NaOH" + "H"_2 What mass of sodium ("Na") is needed to react with 8 grams of water ("H"_2"O")? "(Molar mass of Na" = 23 g//(mol); "molar mass of " "H"_2"O" = 18 g//(mol)")" g_____Na

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zhirou Share
Feb 9, 2018

$\text{10g}$ of sodium. :)

#### Explanation:

This is our balanced equation:

$2 N a + 2 {H}_{2} O \to 2 N a O H + {H}_{2}$

From this, we can tell that for every $1$ mole of water that reacts, exactly $1$ mole of sodium must also react.

$\text{8g}$ of water is:

8/("molar mass of water") = 8/18 = "0.44 moles"

So, this means that $\text{0.44 moles}$ of sodium must also react. Let's convert that to mass:

$0.44 \times \text{molar mass of sodium" = 0.44 xx 23 = "10g}$.

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