# A piece of sodium metal reacts completely with water as follows:

A piece of sodium metal reacts completely with water as follows:
$2Na\left(s\right)+2H2O\left(l\right)⇒2NaOH\left(aq\right)+H2\left(g\right)$
The hydrogen gas generated is collected over water at 25.0 degree C. The volume of the gas is 246 mL measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at 25 degree $C=0.0313$ atm.)
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Daniel Cormack

Step 1
Given information:
The equation for the reaction is
$2Na\left(s\right)+2{H}_{2}O\left(l\right)⇒2NaOH\left(aq\right)+{H}_{2}\left(g\right)$
$-T={25.0}^{\circ }C=\left(273+25\right)K=298K;$
$-V=246mL=0.246L$ is the volume of the gas $\left({H}_{2}\right)$;
$-P-1.00atm$ is the total pressure over the water;
- $R=0.0821Lat\frac{m}{K}mol$ is the ideal gas constant;
- is the vapor pressure of water.
Let's calculate the number of grams of sodium used in the reaction.
Step 2
The total pressure of the mixture is the summary of partial pressures of each component.
${P}_{{}_{\left\{2\right\}}}+{P}_{{H}_{2}O}={P}_{Total}$
First, let's find the partial pressure of the ${H}_{2}$:
${P}_{{}_{\left\{2\right\}}}={P}_{Total}-{P}_{{H}_{2}O}$
${P}_{{H}_{2}}=1.00atm-0.0313atm$
${P}_{{H}_{2}}=0.9687atm$
Now let's calculate the number of moles of ${H}_{2}$. We can use the ideal gas equation.
$PV=nRT$
$n=\frac{PV}{RT}$

$n=9.740\cdot {10}^{-3}mol$
The number of moles of .
Step 3
From the balanced equation for the chemical reaction above, we can conclude that 2mol of Na produces 1mol of ${H}_{2}$.
Therefore, number of moles of Na which produce is:
${n}_{Na}=\frac{2molNa}{1mol{H}_{2}}\cdot 9.740\cdot {10}^{-3}mol{H}_{2}$
${n}_{Na}=0.01948molNa$
Now we can find the mass of sodium in g used in this reaction.
${n}_{Na}=\frac{{m}_{Na}}{{M}_{Na}}$
Therefore,

twineg4
Explanation:
1. Gather all the information in one place
${M}_{r}:22.99$
$2Na+2{H}_{2}O⇒2NaOH+{H}_{2}$
${P}_{\to t}=1.00atm$
${p}_{{H}_{2}O}=0.0313atm$
$T={25.0}^{\circ }C$
$V=246mL$
2. Moles of ${H}_{2}$
To find the moles of hydrogen, we can use the Ideal Gas Law:
$pV=nRT$
(a) Calculate the partial pressure of the hydrogen
${p}_{\to t}={p}_{H2}+{p}_{H2O}$
$1.00atm={p}_{H2}+0.0313atm$
${p}_{H2}=0.9687atm$
(b) Convert the volume to litres
$V=246mL=0.246L$
(c) Convert the temperature to kelvins
$T=\left(25.0+273.15\right)K=298.15K$
(d) Calculate the moles of hydrogen
$0.9687atm×0.246L=n×0.08206L\cdot atm\cdot {K}^{-1}mo{l}^{-1}×298.15K$
$0.2383=24.47nmo{l}^{-1}$
$n=\frac{0.2383}{24.47mo{l}^{-1}}$
$=0.009740mol$
3. Moles of Na
The molar ratio is 2 mol Na: 1 mol ${H}_{2}$
Moles of
4, Mass of Na
Mass of
The mass of Na used was 0.449g.