Answer :
Let's analyze the chemical reaction and given data step by step:
Balanced Equation:
[tex]\[ P_4(s) + 6H_2(g) \rightarrow 4PH_3(g) \][/tex]
This equation tells us the molar ratios between the reactants and the products:
- 1 mole of [tex]\( P_4 \)[/tex] reacts with 6 moles of [tex]\( H_2 \)[/tex] to produce 4 moles of [tex]\( PH_3 \)[/tex].
Given data:
- [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex]
- [tex]\( 3.5 \)[/tex] moles of [tex]\( P_4 \)[/tex]
- [tex]\( 14.0 \)[/tex] moles of [tex]\( PH_3 \)[/tex] produced.
Let's answer the specific parts step by step.
### 1. Moles of [tex]\( P_4 \)[/tex] needed:
How much [tex]\( P_4 \)[/tex] is needed to react with [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex]?
From the balanced equation, [tex]\( 6 \)[/tex] moles of [tex]\( H_2 \)[/tex] react with [tex]\( 1 \)[/tex] mole of [tex]\( P_4 \)[/tex].
[tex]\[ \text{Moles of } P_4 \text{ needed} = \frac{7.0 \, \text{moles of } H_2}{6} = 1.1667 \, \text{moles of } P_4 \][/tex]
So, approximately 1.167 moles of [tex]\( P_4 \)[/tex] are needed to react with 7.0 moles of [tex]\( H_2 \)[/tex].
### 2. Possible Production of [tex]\( PH_3 \)[/tex]:
Given [tex]\( 3.5 \)[/tex] moles of [tex]\( P_4 \)[/tex], how much [tex]\( PH_3 \)[/tex] can be produced?
From the stoichiometry of the balanced equation, [tex]\( 1 \)[/tex] mole of [tex]\( P_4 \)[/tex] produces [tex]\( 4 \)[/tex] moles of [tex]\( PH_3 \)[/tex].
[tex]\[ \text{Moles of } PH_3 \text{ produced} = 3.5 \times 4 = 14.0 \, \text{moles of } PH_3 \][/tex]
Thus, [tex]\( 14.0 \)[/tex] moles of [tex]\( PH_3 \)[/tex] can be produced with the given [tex]\( 3.5 \)[/tex] moles of [tex]\( P_4 \)[/tex].
### 3. Possible Reaction with [tex]\( P_4 \)[/tex]:
Using the [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex], how much [tex]\( P_4 \)[/tex] could react?
From the balanced equation and the molar ratio:
[tex]\[ \text{Moles of } P_4 \text{ that could react} = \frac{7.0 \, \text{moles of } H_2}{\frac{1}{6}} = 7.0 \times 6 = 42.0 \, \text{moles of } P_4 \][/tex]
Thus, [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex] could potentially react with [tex]\( 42.0 \)[/tex] moles of [tex]\( P_4 \)[/tex].
### Summary:
- 1.167 moles of [tex]\( P_4 \)[/tex] are needed to react with [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex].
- [tex]\( 14.0 \)[/tex] moles of [tex]\( PH_3 \)[/tex] can be produced with the given [tex]\( 3.5 \)[/tex] moles of [tex]\( P_4 \)[/tex].
- [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex] could potentially react with [tex]\( 42.0 \)[/tex] moles of [tex]\( P_4 \)[/tex].
Balanced Equation:
[tex]\[ P_4(s) + 6H_2(g) \rightarrow 4PH_3(g) \][/tex]
This equation tells us the molar ratios between the reactants and the products:
- 1 mole of [tex]\( P_4 \)[/tex] reacts with 6 moles of [tex]\( H_2 \)[/tex] to produce 4 moles of [tex]\( PH_3 \)[/tex].
Given data:
- [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex]
- [tex]\( 3.5 \)[/tex] moles of [tex]\( P_4 \)[/tex]
- [tex]\( 14.0 \)[/tex] moles of [tex]\( PH_3 \)[/tex] produced.
Let's answer the specific parts step by step.
### 1. Moles of [tex]\( P_4 \)[/tex] needed:
How much [tex]\( P_4 \)[/tex] is needed to react with [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex]?
From the balanced equation, [tex]\( 6 \)[/tex] moles of [tex]\( H_2 \)[/tex] react with [tex]\( 1 \)[/tex] mole of [tex]\( P_4 \)[/tex].
[tex]\[ \text{Moles of } P_4 \text{ needed} = \frac{7.0 \, \text{moles of } H_2}{6} = 1.1667 \, \text{moles of } P_4 \][/tex]
So, approximately 1.167 moles of [tex]\( P_4 \)[/tex] are needed to react with 7.0 moles of [tex]\( H_2 \)[/tex].
### 2. Possible Production of [tex]\( PH_3 \)[/tex]:
Given [tex]\( 3.5 \)[/tex] moles of [tex]\( P_4 \)[/tex], how much [tex]\( PH_3 \)[/tex] can be produced?
From the stoichiometry of the balanced equation, [tex]\( 1 \)[/tex] mole of [tex]\( P_4 \)[/tex] produces [tex]\( 4 \)[/tex] moles of [tex]\( PH_3 \)[/tex].
[tex]\[ \text{Moles of } PH_3 \text{ produced} = 3.5 \times 4 = 14.0 \, \text{moles of } PH_3 \][/tex]
Thus, [tex]\( 14.0 \)[/tex] moles of [tex]\( PH_3 \)[/tex] can be produced with the given [tex]\( 3.5 \)[/tex] moles of [tex]\( P_4 \)[/tex].
### 3. Possible Reaction with [tex]\( P_4 \)[/tex]:
Using the [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex], how much [tex]\( P_4 \)[/tex] could react?
From the balanced equation and the molar ratio:
[tex]\[ \text{Moles of } P_4 \text{ that could react} = \frac{7.0 \, \text{moles of } H_2}{\frac{1}{6}} = 7.0 \times 6 = 42.0 \, \text{moles of } P_4 \][/tex]
Thus, [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex] could potentially react with [tex]\( 42.0 \)[/tex] moles of [tex]\( P_4 \)[/tex].
### Summary:
- 1.167 moles of [tex]\( P_4 \)[/tex] are needed to react with [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex].
- [tex]\( 14.0 \)[/tex] moles of [tex]\( PH_3 \)[/tex] can be produced with the given [tex]\( 3.5 \)[/tex] moles of [tex]\( P_4 \)[/tex].
- [tex]\( 7.0 \)[/tex] moles of [tex]\( H_2 \)[/tex] could potentially react with [tex]\( 42.0 \)[/tex] moles of [tex]\( P_4 \)[/tex].
