already missed question so I just need help figuring out what I did. Small amounts of chlorine gas can be gener-
ated in the laboratory from the reaction of
manganese(IV) oxide with hydrochloric acid:
4 HCl(aq) + MnO
2
(s)
→
2 H
2
O(
ℓ
) + MnCl
2
(s) + Cl
2
(g).
What mass of Cl
2
can be produced from 38
.
1 g
of MnO
2
with an excess of HCl(aq)?
Answer in units of g.
010 (part 2 of 3) 10.0 points
What volume of chlorine gas (of density 3.17
g/L) will be produced from the reaction of
381 mL of 0
.
36 M HCl(aq) with an excess of
MnO
2
?
Answer in units of L.
011 (part 3 of 3) 10.0 points
Suppose only 116 mL of chlorine gas was pro-
duced in the reaction in part two. What is
the percentage yield of the reaction?
Answer in units of %. I think the answer to part 3 is 1.4%. I just need to know if I finally figured out how to do it.
To determine the answers to the given questions, we need to use stoichiometry and conversion factors. Let's break down each question step by step:
1. What mass of Cl2 can be produced from 38.1 g of MnO2 with an excess of HCl(aq)?
First, calculate the molar mass of MnO2:
Molar mass of MnO2 = (1 × atomic mass of Mn) + (2 × atomic mass of O)
Next, convert the mass of MnO2 to moles using the molar mass:
Moles of MnO2 = Mass of MnO2 / Molar mass of MnO2
Using the balanced equation, we know that the ratio of MnO2 to Cl2 is 1:1. Therefore, the moles of Cl2 produced will be equal to the moles of MnO2.
Finally, calculate the mass of Cl2 produced using the molar mass of Cl2:
Mass of Cl2 = Moles of Cl2 × Molar mass of Cl2
2. What volume of chlorine gas (of density 3.17 g/L) will be produced from the reaction of 381 mL of 0.36 M HCl(aq) with an excess of MnO2?
First, calculate the number of moles of HCl:
Moles of HCl = Volume of HCl(aq) × Molarity of HCl
Using the balanced equation, we know that the ratio of HCl to Cl2 is 4:1. Therefore, the moles of Cl2 produced will be one-fourth the moles of HCl.
Next, convert moles of Cl2 to volume using the ideal gas law:
Volume of Cl2 = Moles of Cl2 × Molar volume of gas at STP
Since the density of Cl2 is given, you can also use this information to calculate the volume of Cl2:
Volume of Cl2 = Mass of Cl2 / Density of Cl2
3. Suppose only 116 mL of chlorine gas was produced in the reaction. What is the percentage yield of the reaction?
To calculate the percentage yield, we need to compare the actual yield (116 mL) with the theoretical yield (calculated in question 2). The percentage yield is given by:
Percentage yield = (Actual yield / Theoretical yield) × 100
You mentioned that the answer to part 3 is 1.4%. To verify if this is correct, perform the calculations and check if the resulting percentage matches.
Remember to follow your calculations step by step and ensure that the conversion factors and units are consistent throughout.