Na + H2O ? NaOH + H2
a). Balance the equation (show all work to get full credit)
b). How many moles of sodium are required to produce 5.0 moles of NaOH, sodium hydroxide?
c). How many moles of H2 are produced in the reaction with 5.0 grams of water?
The Correct Answer and Explanation is :
Let’s break this down:
a) Balance the equation:
The unbalanced chemical reaction is:
[
\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2
]
To balance the equation, follow these steps:
- Balance Sodium (Na): Each Na on the left side reacts to form NaOH on the right side. Thus, we need one Na atom for each NaOH molecule:
[
\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2
] - Balance Hydrogen (H): On the right side, we have 1 molecule of H2 and 1 molecule of NaOH, which gives 2 hydrogens. On the left, we have H2O (which contains 2 hydrogens). This is already balanced.
- Balance Oxygen (O): We have 1 oxygen atom in NaOH on the right side, and 1 in H2O on the left side. This is balanced.
The equation is already balanced as:
[
\text{2 Na} + \text{2 H}_2\text{O} \rightarrow \text{2 NaOH} + \text{H}_2
]
b) How many moles of sodium are required to produce 5.0 moles of NaOH?
The balanced equation shows that 2 moles of sodium (Na) react with 2 moles of water (H2O) to produce 2 moles of sodium hydroxide (NaOH). Therefore, the number of moles of sodium required to produce NaOH is directly proportional to NaOH.
[
\text{Moles of Na} = \frac{2 \, \text{mol Na}}{2 \, \text{mol NaOH}} \times 5.0 \, \text{mol NaOH} = 5.0 \, \text{mol Na}
]
Thus, 5.0 moles of sodium are required to produce 5.0 moles of NaOH.
c) How many moles of H2 are produced in the reaction with 5.0 grams of water?
First, convert 5.0 grams of water (H2O) to moles:
- The molar mass of water (H2O) is 18.015 g/mol.
[
\text{Moles of H}_2\text{O} = \frac{5.0 \, \text{g H}_2\text{O}}{18.015 \, \text{g/mol}} = 0.2775 \, \text{mol H}_2\text{O}
]
From the balanced equation:
[
2 \, \text{mol Na} + 2 \, \text{mol H}_2\text{O} \rightarrow 2 \, \text{mol NaOH} + \text{H}_2
]
This shows that 2 moles of water produce 1 mole of hydrogen gas (H2). So, the moles of H2 produced will be:
[
\text{Moles of H}_2 = \frac{1 \, \text{mol H}_2}{2 \, \text{mol H}_2\text{O}} \times 0.2775 \, \text{mol H}_2\text{O} = 0.1388 \, \text{mol H}_2
]
Therefore, 0.1388 moles of hydrogen gas are produced when 5.0 grams of water react in the reaction.
Explanation:
In the reaction, sodium reacts with water to produce sodium hydroxide and hydrogen gas. This process is a single displacement reaction. The coefficients in the balanced equation indicate the stoichiometric relationship between the reactants and products. For every 2 moles of water reacted, 1 mole of hydrogen gas is produced, which allows us to determine the amount of hydrogen gas produced from a given mass of water. Similarly, the amount of sodium required to produce a certain amount of sodium hydroxide can be calculated using the molar ratios derived from the balanced equation.