Question
When the alkali metal rubidium reacts with water, it forms rubidium hydroxide and hydrogen gas. Use what you've learned about balancing equations to select the correct coefficients to represent the law of conservation of mass for the reaction below.
rubidium + water → rubidium hydroxide + hydrogen
Rb (s) + H2O (l) → RbOH (aq) +H2 (g)
(4 points)
Rb (s) +
H2O (l)
→
RbOH (aq) +
H2 (g)
1, 2, 3, 4
rubidium + water → rubidium hydroxide + hydrogen
Rb (s) + H2O (l) → RbOH (aq) +H2 (g)
(4 points)
Rb (s) +
H2O (l)
→
RbOH (aq) +
H2 (g)
1, 2, 3, 4
Answers
GPT-4o mini
To balance the chemical equation for the reaction of rubidium with water, we need to ensure that the number of atoms for each element is the same on both sides of the equation.
The unbalanced equation is:
\[ \text{Rb} (s) + \text{H}_2\text{O} (l) \rightarrow \text{RbOH} (aq) + \text{H}_2 (g) \]
1. **Count the atoms on both sides:**
- **Reactants:**
- Rb: 1
- H: 2 (from H₂O)
- O: 1
- **Products:**
- Rb: 1
- H: 3 (1 from RbOH and 2 from H₂)
- O: 1
2. **Balancing the hydrogen atoms:**
We have 2 hydrogen atoms from water and only 3 hydrogen atoms total in products (1 from RbOH and 2 from H₂). Thus, we need to balance the number of hydrogen atoms.
We can change the coefficient of water (H₂O) to 2:
\[ \text{Rb} + 2 \text{H}_2\text{O} \rightarrow \text{RbOH} + \text{H}_2 \]
3. **Now recount the atoms:**
- **New Reactants:**
- Rb: 1
- H: 4 (2 from each of the 2 water molecules)
- O: 2
- **Products:**
- Rb: 1
- H: 3 (1 from RbOH and 2 from H₂)
- O: 1
4. **To fully balance the equation:**
We need to add another rubidium hydroxide:
\[ 2 \text{Rb} + 2 \text{H}_2\text{O} \rightarrow 2 \text{RbOH} + \text{H}_2 \]
Now, let's recount the atoms again:
- **Reactants:**
- Rb: 2
- H: 4 (2 from each of the 2 water molecules)
- O: 2 (1 from each of the 2 water molecules)
- **Products:**
- Rb: 2 (from 2 RbOH)
- H: 4 (2 from 2 RbOH and 2 from H₂)
- O: 2 (2 from the 2 RbOH)
Now both sides are balanced.
Thus, the correct coefficients for the reaction are:
**2 Rb (s) + 2 H₂O (l) → 2 RbOH (aq) + H₂ (g)**
In your case, using the substitutions given,
1. Rb (s) + **2** H₂O (l) → **2** RbOH (aq) + H₂ (g)
So the responses to your answer are:
**1, 2, 2, 1** (for Rb, H2O, RbOH, and H2 respectively).
The unbalanced equation is:
\[ \text{Rb} (s) + \text{H}_2\text{O} (l) \rightarrow \text{RbOH} (aq) + \text{H}_2 (g) \]
1. **Count the atoms on both sides:**
- **Reactants:**
- Rb: 1
- H: 2 (from H₂O)
- O: 1
- **Products:**
- Rb: 1
- H: 3 (1 from RbOH and 2 from H₂)
- O: 1
2. **Balancing the hydrogen atoms:**
We have 2 hydrogen atoms from water and only 3 hydrogen atoms total in products (1 from RbOH and 2 from H₂). Thus, we need to balance the number of hydrogen atoms.
We can change the coefficient of water (H₂O) to 2:
\[ \text{Rb} + 2 \text{H}_2\text{O} \rightarrow \text{RbOH} + \text{H}_2 \]
3. **Now recount the atoms:**
- **New Reactants:**
- Rb: 1
- H: 4 (2 from each of the 2 water molecules)
- O: 2
- **Products:**
- Rb: 1
- H: 3 (1 from RbOH and 2 from H₂)
- O: 1
4. **To fully balance the equation:**
We need to add another rubidium hydroxide:
\[ 2 \text{Rb} + 2 \text{H}_2\text{O} \rightarrow 2 \text{RbOH} + \text{H}_2 \]
Now, let's recount the atoms again:
- **Reactants:**
- Rb: 2
- H: 4 (2 from each of the 2 water molecules)
- O: 2 (1 from each of the 2 water molecules)
- **Products:**
- Rb: 2 (from 2 RbOH)
- H: 4 (2 from 2 RbOH and 2 from H₂)
- O: 2 (2 from the 2 RbOH)
Now both sides are balanced.
Thus, the correct coefficients for the reaction are:
**2 Rb (s) + 2 H₂O (l) → 2 RbOH (aq) + H₂ (g)**
In your case, using the substitutions given,
1. Rb (s) + **2** H₂O (l) → **2** RbOH (aq) + H₂ (g)
So the responses to your answer are:
**1, 2, 2, 1** (for Rb, H2O, RbOH, and H2 respectively).