Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
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].
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
