How many kilojoules in 100 grams

What is the theoretical yield of methanol (CH3OH) when 12.0 grams of H2 is mixed with 74.5 grams of CO? CO + 2H2 CH3OH

AlladinOne [14]

1) We need to convert 12.0 g of H2 into moles of H2, and 74.5 grams of CO into moles of CO
Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol

12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO

2) Now we can use reaction to find out what substance will react completely, and what will be leftover.

CO + 2H2 -------> CH3OH
 1 mol 2 mol
given 2.66 mol 6 mol (excess)

How much
we need CO? 3 mol 6 mol

We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.

3) CO + 2H2 -------> CH3OH
 1 mol 1 mol
 2.66 mol 2.66 mol

4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol

2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH

Answer is D) 85.12 grams.

3 0

3 years ago

Which of the following would be an example of basic research?

Nitella [24]

B) Morrison and Franscioni's research done to create the Frisbee.

5 0

3 years ago

Read 2 more answers

The combustion of ethane ( C 2 H 6 ) (C2H6) produces carbon dioxide and steam. 2 C 2 H 6 ( g ) + 7 O 2 ( g ) ⟶ 4 CO 2 ( g ) + 6

REY [17]

Answer:

There is a production of 11.6 moles of CO₂

Explanation:

The reaction is this:

2C₂H₆(g) + 7O₂(g) ⟶ 4CO₂(g) + 6H₂O(g)

2 moles of ethane reacts with 7 moles of oxygen, to make 4 mol of dioxide and 6 moles of water vapor.

If the oxygen is in excess, we make the calculate with the ethane (limiting reactant)

2 moles of ethane produce 4 moles of dioxide

5.8 moles of ethane produce (5.8 .4)/2 = 11.6 moles

5 0

3 years ago

Write a complete, balanced chemical equation where tin metal reacts with aqueous hydrochloric acid to produce tin(II) chloride a

AleksAgata [21]

Answer:

1. The balanced equation is given below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

2a. H is oxidized.

2b. Sn is reduced.

Explanation:

1. Balanced equation for the reaction between tin (Sn) metal and aqueous hydrochloric acid (HCl) to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂).

This is illustrated below:

Sn (s) + HCl (aq) –> SnCl₂ (aq) + H₂ (g)

There are 2 atoms of Cl on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

Now, the equation is balanced

2. Determination of the element that is oxidize and reduced.

This can be obtained as follow:

We shall determine the change in oxidation number of each element.

NOTE:

a. The oxidation number of H is always +1 except in hydrides where it is –1.

b. The oxidation state of Cl is always –1.

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

For Tin (Sn):

Sn = 0

SnCl₂ = 0

Sn + 2Cl = 0

Cl = – 1

Sn + 2(–1) = 0

Sn – 2 = 0

Collect like terms

Sn = 0 + 2

Sn = +2

Therefore, the oxidation number of Tin (Sn) changes from 0 to +2

For H:

H = +1

H₂ = 0

The oxidation number of H changes from +1 to 0

For Cl:

Cl is always –1. Therefore no change.

Summary:

Element >>Change in oxidation number

Sn >>>>>>>From 0 to +2

H >>>>>>>>From +1 to 0

Cl >>>>>>>No change

Therefore,

Sn is reduced since its oxidation number increased from 0 to +2.

H is oxidized since it oxidation number reduced from +1 to 0

4 0

3 years ago

How are oxidation-reduction reactions related to how you use energy?

Tems11 [23]

Answer:

In oxidation reduction reactions, one species gets reduced by taking on electron(s) and another species gets oxidized by losing electrons. They also flow by a wire

Explanation:

8 0

3 years ago