Question

For the following reaction, 6.80 grams of hydrochloric acid are mixed with excess iron. The reaction yields 8.51 grams of iron(II) chloride. iron (s) + hydrochloric acid (aq) rightarrow iron(II) chloride (aq) + hydrogen (g) a.What is the theoretical yield of iron(II) chloride? grams b. What is the percent yield for this reaction?

Answer 1

Answer:

A. 11.83g

B. 71.9%

Explanation:

We'll begin by writing the balanced for the reaction. This is shown below:

Fe + 2HCl —> FeCl2 + H2

A. Step 1

Determination of the mass of HCl that reacted and the mass of FeCl2 produced from the balanced equation. This is illustrated below:

Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 2 x 36.5 = 73g

Molar Mass of FeCl2 = 56 + (35.5x2) = 56 + 71 = 127g

Summary:

From the balanced equation above,

73g of HCl produced 127g of FeCl2.

A. Step 2:

Determination of the theoretical yield of FeCl2.

The theoretical yield of FeCl2 can be obtained as follow:

Fe + 2HCl —> FeCl2 + H2

From the balanced equation above,

73g of HCl produced 127g of FeCl2.

Therefore, 6.80g of HCl will produce = (6.80 x 127)/73 = 11.83g of FeCl2.

Therefore, the theoretical yield of FeCl2 is 11.83g

B. Determination of the percentage yield.

Actual yield = 8.51g

Theoretical yield = 11.83g

Percentage yield =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 8.51/11.83 x100

Percentage yield = 71.9%

Therefore, the percentage yield of FeCl2 is 71.9%

Answer 2

Answer:

a. $m_{FeCl_2}^{theoretical}=11.82g$

b. $Y=72.0\%$

Explanation:

Hello,

a. In this case, the undergoing chemical reaction turns out:

$Fe(s) + 2HCl (aq) \rightarrow FeCl_2 (aq) + H_2(g)$

Hence, for the initial grams of 6.80 g of HCl, we compute the theoretical yield of iron (II) chloride by applying the following stoichiometric factor with the 2:1 mole ratio between them:

$m_{FeCl_2}^{theoretical}=6.80gHCl*\frac{1molHCl}{36.45gHCl}*\frac{1molFeCl_2}{2molHCl}*\frac{126.75gFeCl_2}{1molFeCl_2} \\m_{FeCl_2}^{theoretical}=11.82g$

b. Now, we use the percent yield formula as:

$Y=\frac{m_{FeCl_2}^{obtained}}{m_{FeCl_2}^{theoretical}} *100\%=\frac{8.51g}{11.82g}*100\%\\\\Y=72.0\%$

Best regards.