Q1) 5 liters of Butane gas (C4H10) is reacted with plenty of oxygen at 27 oC and 1 atm to produce carbon dioxide and water vapor. Calculate the following:

1- The volume of CO2 produced at 1 atm and 400 oC.
2- The mass of H2O vapor produced.
3- The mol % of the products.
Given:
C4H10 + O2 = CO2 + H2O (Unbalanced equation)

Q2) 10 M of hydrochloric acid (HCl) is reacted with zinc. If 2 liters of HCl is reacted with the stoichiometric amount of Zn. Calculate the mass of zinc required and the volume of hydrogen evolved during this reaction if it was measured at 1 atm and 27 oC. What is the mass of zinc chloride produced?
Zn + HCl = ZnCl2 + H2 (g) (Unbalanced equation)

To get the answers to these questions, we need to balance the given chemical equations first. Once we have balanced equations, we can use stoichiometry to calculate the desired quantities.

Q1) Balancing the equation:
C4H10 + 13/2 O2 = 4 CO2 + 5 H2O

1) To find the volume of CO2 produced at 1 atm and 400 oC, we need to use the ideal gas law equation:
PV = nRT

First, we need to calculate the number of moles of CO2:
Since CO2 is a product of the reaction with a coefficient of 4, the number of moles of CO2 produced will be equal to the number of moles of C4H10 used in the reaction.

Number of moles of CO2 = Number of moles of C4H10 = 5 liters / 22.4 liters/mol (at STP)
Number of moles of CO2 = 0.2232 moles

Now we can use the ideal gas law:
PV = nRT

P = 1 atm
V = volume of CO2 (to be calculated)
n = 0.2232 moles
R = 0.0821 L*atm/(mol*K)
T = 400 oC = 400 + 273 = 673 K

Solving for V:
V = (nRT) / P
V = (0.2232 moles * 0.0821 L*atm/(mol*K) * 673 K) / 1 atm

V ≈ 11.69 liters

Therefore, the volume of CO2 produced at 1 atm and 400 oC is approximately 11.69 liters.

2) To calculate the mass of H2O vapor produced, we need to use the molar mass of water (H2O) and the number of moles of water produced in the reaction.

Molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol

Number of moles of H2O = Number of moles of C4H10 used in the reaction (since H2O is a product)
Number of moles of H2O = 0.2232 moles

Mass of H2O = number of moles of H2O * molar mass of H2O
Mass of H2O = 0.2232 moles * 18.02 g/mol

Mass of H2O = 4.02 grams

Therefore, the mass of H2O vapor produced is 4.02 grams.

3) To calculate the mol % of the products, we need to find the number of moles of CO2 and H2O and calculate their respective percentages by moles.

Number of moles of CO2 = Number of moles of C4H10 = 0.2232 moles (as calculated earlier)
Number of moles of H2O = 0.2232 moles (since it is a product)

Total number of moles of products = Number of moles of CO2 + Number of moles of H2O
Total number of moles of products = 0.2232 moles + 0.2232 moles = 0.4464 moles

Mol % of CO2 = (Number of moles of CO2 / Total number of moles of products) * 100
Mol % of CO2 = (0.2232 moles / 0.4464 moles) * 100

Mol % of CO2 ≈ 50%

Mol % of H2O = (Number of moles of H2O / Total number of moles of products) * 100
Mol % of H2O = (0.2232 moles / 0.4464 moles) * 100

Mol % of H2O ≈ 50%

Therefore, the mol % of both CO2 and H2O in the products is approximately 50% each.

Q2) Balancing the equation:
Zn + 2 HCl = ZnCl2 + H2 (g)

To calculate the mass of zinc required and the volume of hydrogen evolved, we need to use stoichiometry.

First, we need to calculate the number of moles of HCl that reacted:
Number of moles of HCl = Molarity * volume (in liters)
Number of moles of HCl = 10 M * 2 L = 20 moles

From the balanced equation, we see that the ratio of Zn to HCl is 1:2. This means that for every mole of Zn, 2 moles of HCl react.

Therefore, for 20 moles of HCl, we need:

Number of moles of Zn = 20 moles / 2 = 10 moles

To calculate the mass of Zn required, we need to use the molar mass of Zn:
Molar mass of Zn = 65.38 g/mol

Mass of Zn = number of moles of Zn * molar mass of Zn
Mass of Zn = 10 moles * 65.38 g/mol

Mass of Zn = 653.8 grams

Therefore, the mass of zinc required is 653.8 grams.

To calculate the volume of hydrogen evolved, we can use the ideal gas law equation:
PV = nRT

P = 1 atm
V = volume of H2 (to be calculated)
n = number of moles of H2 (from the balanced equation, it is the same as the number of moles of Zn)
R = 0.0821 L*atm/(mol*K)
T = 27 oC = 27 + 273 = 300 K

Solving for V:
V = (nRT) / P
V = (10 moles * 0.0821 L*atm/(mol*K) * 300 K) / 1 atm

V ≈ 24.63 liters

Therefore, the volume of hydrogen evolved during this reaction, measured at 1 atm and 27 oC, is approximately 24.63 liters.

To calculate the mass of zinc chloride produced, we need to use stoichiometry. From the balanced equation, we see that the ratio of Zn to ZnCl2 is 1:1, meaning that the number of moles of Zn is equal to the number of moles of ZnCl2.

Number of moles of ZnCl2 = Number of moles of Zn = 10 moles

To calculate the mass of ZnCl2, we need to use the molar mass of ZnCl2:
Molar mass of ZnCl2 = 65.38 g/mol (for Zn) + 2 * (35.45 g/mol) (for Cl)

Molar mass of ZnCl2 = 136.28 g/mol

Mass of ZnCl2 = number of moles of ZnCl2 * molar mass of ZnCl2
Mass of ZnCl2 = 10 moles * 136.28 g/mol

Mass of ZnCl2 = 1362.8 grams

Therefore, the mass of zinc chloride produced is 1362.8 grams.