Answer to Question #288102 in Organic Chemistry for Buddy

Question

1. When 9.0 g of hydrogen gas, H₂, reacts with oxygen gas, O₂, 73.0 g of water is produced.    2H2 + O2 --> 2H2O a) What is the theoretical yield of water? b) What is the percentage yield?

2. Powdered zinc metal reacts with sulphur (S₈) when heated to produce zinc sulphide.

a) What mass of product can be produced with 25 g of zinc?

b) What is the actual yield if the percentage yield was 80%?

3. The thermite reaction has been used to weld railroad rails, make bombs, and ignite solid rocket fuel. The equation of the reaction is:

a) What is the theoretical yield of iron if 100.0 g of Fe₂O₃ is used?

b) What is the percentage yield if 52.3 g of iron is actually produced?

Answer 1

(1)

The equation for this reaction is

2H₂ + O₂ = 2H₂O;

Having a mass of hydrogen we can find the quantity substance

n(H₂) = m(H₂)/M(H₂);

M(H₂) = 1 *2 = 2 g/mol

n(H₂) = 9 g / 2 g/mol = 4.5 moles;

The amount of substance of hydrogen with the amount of substance of water as

a 2 to 2 or 1 to 1.

so n(H₂O) = 4.5 moles

Theoretical mass of water will be

n(H₂O) = m(H₂O)/M(H₂O);

M(H₂O) = 1 *2 + 16 = 18 g/mol

m (H₂O) 4.5 moles * 18 g/mol = 81 g.

so the theoretical yield of water is 81 g.

The percentage yield will be

m(H₂O)practical / m(H₂O)theoretical = 73.0 g / 81.0 g = 0.9012 or 90.12 %

(2)

The equation of the reaction is:

Zn+S =ZnS

Begin by determining the molar mass of each compound involved in the reaction. Using atomic masses from the periodic table, we will find the following:

M(ZnS) = 97.43 g/mol;

Ar(Zn) = 65.37 g/mol.

a) The mass of product determine from the equation of the reaction:

Zn + S= ZnS

m(ZnS)= $\frac{97.43 \times 25}{65.37}=37.26 g$

b) The actyal yield of ZnS is:

m(ZnS)=37.26×0.8=29.81 g. Answer:

1. a) The theoretical mass of ZnS is 37.26 g;
2. b) The actual yield of ZnS is: 29.81 g. 

(3)

The equation of the reaction is:

Fe₂O₃(s) + 2Al(s) → 2Fe(l) + Al₂O₃(s) ... balanced equation

Begin by determining the molar mass of each compound involved in the reaction.

Using atomic masses from the periodic table, we will find the following

$M(Fe_2O_3)=159.7g/mol;\\Ar(Fe)=55.85 g/mol.$

a)

The theoretical yield of iron determine from the equation of the reaction:

$Fe_2O_3+2Al=2\space Fe +Al_2O_3\\m(Fe)=\frac{100 \times 2\times 55.85}{159.7}=69.9g$

b) The percentage yield of iron is:

n=$\frac{52.3}{69.9}\times 100=74.8\%$

a) m(Fe)=69.9 g;

b) yield of iron is 74.8%