Question

the chemical equation for the reaction of sodium metal with water is 2na(s) + 2h₂o(l) → h₂(g) + 2naoh(aq). the reaction of 23 g of sodium with 9 g of water produces 1 g of hydrogen gas. what mass of sodium hydroxide is produced? a) 32 g b) 31 g c) 30 g d) 33 g

Explanation:

Step1: Write the balanced chemical equation

$2Na(s)+2H_2O(l)
ightarrow H_2(g) + 2NaOH(aq)$

Step2: Calculate the molar - masses

The molar mass of $H_2$ is $M_{H_2}=2\times1\ g/mol = 2\ g/mol$, and the molar mass of $NaOH$ is $M_{NaOH}=23 + 16+1=40\ g/mol$.

Step3: Use the mole - ratio from the balanced equation

From the balanced equation, the mole - ratio of $H_2$ to $NaOH$ is $n_{H_2}:n_{NaOH}=1:2$.
We know the mass of $H_2$ produced is $m_{H_2} = 1\ g$. The number of moles of $H_2$, $n_{H_2}=\frac{m_{H_2}}{M_{H_2}}=\frac{1\ g}{2\ g/mol}=0.5\ mol$.
Since $n_{H_2}:n_{NaOH}=1:2$, the number of moles of $NaOH$, $n_{NaOH}=2n_{H_2}=2\times0.5\ mol = 1\ mol$.

Step4: Calculate the mass of $NaOH$

The mass of $NaOH$, $m_{NaOH}=n_{NaOH}\times M_{NaOH}=1\ mol\times40\ g/mol = 40\ g$. But we can also use the law of conservation of mass.
The total mass of reactants is $m_{Na}+m_{H_2O}=23\ g + 9\ g=32\ g$. The mass of $H_2$ produced is $1\ g$.
According to the law of conservation of mass ($m_{reactants}=m_{products}$), if we let the mass of $NaOH$ be $x$, then $23 + 9=1 + x$.
Solving for $x$ gives $x=(23 + 9)-1=31\ g$.

Answer:

B) 31 g