Question

co + 2 h₂ → ch₃oh

1. if you insert 2.27 moles of co, how many moles of h₂ are also used?
2. if you insert 0.75 moles of h₂, how many moles of ch₃oh are produced?

2 fe₂o₃ + 3 c → 4 fe + 3 co₂

1. how many moles of fe will be produced if 0.897 moles of co₂ are produced?
2. how many moles of fe₂o₃ are required if 1.33 moles of fe are produced?

Explanation:

Step1: Analyze the reaction equation

For the reaction \(CO + 2H_{2}\to CH_{3}OH\), the mole ratio of \(CO\) to \(H_{2}\) is \(1:2\).

Step2: Calculate moles of \(H_{2}\)

Given \(n(CO)=2.27\) mol. Using the mole ratio \(\frac{n(H_{2})}{n(CO)}=\frac{2}{1}\), so \(n(H_{2}) = 2\times n(CO)\).
Substitute \(n(CO) = 2.27\) mol: \(n(H_{2})=2\times2.27 = 4.54\) mol.

Step3: Analyze the second - part of the first reaction

For \(CO + 2H_{2}\to CH_{3}OH\), the mole ratio of \(H_{2}\) to \(CH_{3}OH\) is \(2:1\). Given \(n(H_{2}) = 0.75\) mol. Using the mole ratio \(\frac{n(CH_{3}OH)}{n(H_{2})}=\frac{1}{2}\), so \(n(CH_{3}OH)=\frac{1}{2}\times n(H_{2})\).
Substitute \(n(H_{2}) = 0.75\) mol: \(n(CH_{3}OH)=\frac{0.75}{2}=0.375\) mol.

Step4: Analyze the second reaction \(2Fe_{2}O_{3}+3C\to4Fe + 3CO_{2}\)

The mole ratio of \(Fe\) to \(CO_{2}\) is \(\frac{4}{3}\). Given \(n(CO_{2}) = 0.897\) mol. Using the mole ratio \(\frac{n(Fe)}{n(CO_{2})}=\frac{4}{3}\), so \(n(Fe)=\frac{4}{3}\times n(CO_{2})\).
Substitute \(n(CO_{2}) = 0.897\) mol: \(n(Fe)=\frac{4\times0.897}{3}=1.196\) mol.

Step5: Analyze the mole ratio for \(Fe_{2}O_{3}\) and \(Fe\)

For \(2Fe_{2}O_{3}+3C\to4Fe + 3CO_{2}\), the mole ratio of \(Fe_{2}O_{3}\) to \(Fe\) is \(\frac{2}{4}=\frac{1}{2}\). Given \(n(Fe) = 1.33\) mol. Using the mole ratio \(\frac{n(Fe_{2}O_{3})}{n(Fe)}=\frac{1}{2}\), so \(n(Fe_{2}O_{3})=\frac{1}{2}\times n(Fe)\).
Substitute \(n(Fe) = 1.33\) mol: \(n(Fe_{2}O_{3})=\frac{1.33}{2}=0.665\) mol.

Answer:

1. \(4.54\) moles of \(H_{2}\) are used.
2. \(0.375\) moles of \(CH_{3}OH\) are produced.
3. \(1.196\) moles of \(Fe\) are produced.
4. \(0.665\) moles of \(Fe_{2}O_{3}\) are required.