Question

use the references to access important values if needed for this question. for the following reaction, 8.16 grams of sodium are allowed to react with 9.41 grams of water. sodium (s) + water (l) → sodium hydroxide (aq) + hydrogen (g) what is the maximum amount of sodium hydroxide that can be formed? mass = g what is the formula for the limiting reactant? what amount of the excess reactant remains after the reaction is complete? mass = g submit answer retry entire group 2 more group attempts remaining write the balanced equation.

Explanation:

Step1: Balance the chemical equation

The balanced equation is \(2Na(s)+2H_{2}O(l)
ightarrow 2NaOH(aq)+H_{2}(g)\)

Step2: Calculate the moles of reactants

• Molar mass of \(Na\): \(M_{Na}=22.99\ g/mol\). Moles of \(Na\), \(n_{Na}=\frac{m_{Na}}{M_{Na}}=\frac{8.16\ g}{22.99\ g/mol}\approx0.355\ mol\)
• Molar mass of \(H_{2}O\): \(M_{H_{2}O} = 18.02\ g/mol\). Moles of \(H_{2}O\), \(n_{H_{2}O}=\frac{m_{H_{2}O}}{M_{H_{2}O}}=\frac{9.41\ g}{18.02\ g/mol}\approx0.522\ mol\)

Step3: Determine the limiting reactant

From the balanced equation, the mole ratio of \(Na\) to \(H_{2}O\) is \(1:1\).
For \(Na\): If all \(Na\) reacts (\(n_{Na} = 0.355\ mol\)), the moles of \(H_{2}O\) required is \(0.355\ mol\). Since \(0.355\ mol<0.522\ mol\) (moles of \(H_{2}O\) available), \(Na\) is the limiting reactant.

Step4: Calculate the moles of \(NaOH\) formed

From the balanced equation, mole ratio of \(Na\) to \(NaOH\) is \(1:1\). So, \(n_{NaOH}=n_{Na} = 0.355\ mol\)

• Molar mass of \(NaOH\): \(M_{NaOH}=40.00\ g/mol\). Mass of \(NaOH\), \(m_{NaOH}=n_{NaOH}\times M_{NaOH}=0.355\ mol\times40.00\ g/mol = 14.2\ g\)

Step5: Calculate the moles of excess reactant (\(H_{2}O\)) remaining

Moles of \(H_{2}O\) reacted \(= 0.355\ mol\)
Moles of \(H_{2}O\) remaining \(n_{H_{2}O,\text{remaining}}=n_{H_{2}O,\text{initial}}-n_{H_{2}O,\text{reacted}}=0.522\ mol - 0.355\ mol=0.167\ mol\)

• Mass of \(H_{2}O\) remaining \(m_{H_{2}O,\text{remaining}}=n_{H_{2}O,\text{remaining}}\times M_{H_{2}O}=0.167\ mol\times18.02\ g/mol\approx3.01\ g\)

Answer:

Mass of \(NaOH = 14.2\ g\)
Formula for the limiting reactant: \(Na\)
Mass of excess reactant (\(H_{2}O\)) remaining \(=3.01\ g\)