20 5.5 g of bronze (an alloy of copper and tin, Cu-Sn) containing 23 % of Sn are dipped in an excess of HCI. a) Write the balanced equation of the occur- ring reaction. b) What is the volume of H2 evolved when the molar volume of the gas is 24 L.mol? c) What is the final concentration of Snt ions produced if the volume of the solution is 200 mL?
a. Sn + 2HCl ---> SnCl2 + H2
b. 5.5 x 0.23 = 1.27 g Sn
moles Sn = g/atomic mass = 1.27/atomic mass Sn = ?. Look up the atomic mass Sn.
mols H2 produced = mols Sn
volume H2 = molsSn x 24 L/mol = ?
c. You will have moles Sn = 1.27 g Sn/atomic mass Sn = ? from above.
The M = moles/L solution you will have ?mols/0.200 L = ?M
a) Well, bronze is an alloy of copper and tin, so let's start with the balanced equation:
Cu-Sn + 2HCl → SnCl2 + CuCl2
b) To calculate the volume of H2 evolved, we need to find the number of moles of Sn in the bronze. We know that the bronze contains 23% tin, so we can calculate the mass of tin:
5.5 g * 0.23 = 1.265 g of Sn
Now, we need to convert grams of Sn to moles of Sn using the molar mass of Sn:
1.265 g Sn * (1 mole Sn / 118.71 g Sn) = 0.01067 moles of Sn
According to the balanced equation, 1 mole of Sn will produce 1 mole of H2. Since the molar volume of gas (at STP) is 24 L/mol, we can find the volume of H2:
0.01067 moles of Sn * 24 L/mol = 0.256 L
So, the volume of H2 gas evolved is approximately 0.256 liters.
c) We need to find the concentration of Sn ions in the final solution. Let's use the equation from part a) to determine the stoichiometry of SnCl2 in the reaction:
Cu-Sn + 2HCl → SnCl2 + CuCl2
From the equation, we can see that 1 mole of SnCl2 is produced for every mole of Cu-Sn. Since we have 0.01067 moles of Sn, that means 0.01067 moles of SnCl2 are produced.
To find the concentration, we can use the equation:
C = n/V
Where C is the concentration (in moles per liter), n is the number of moles, and V is the volume (in liters). Given that the volume is 200 mL, we'll convert it to liters by dividing by 1000:
V = 200 mL / 1000 = 0.2 L
Now, we can substitute the values into the equation:
C = 0.01067 moles / 0.2 L = 0.05335 moles/L
So, the final concentration of Sn ions produced is approximately 0.053 moles per liter.
Hope that helps! Let me know if you have any other questions.
a) The balanced equation for the reaction between bronze and hydrochloric acid (HCl) can be written as follows:
Sn + 2HCl → SnCl2 + H2
b) To determine the volume of H2 gas evolved, we need to calculate the number of moles of Sn present in 5.5 g of bronze and then use this value to calculate the number of moles of H2 gas using stoichiometry.
First, we need to calculate the number of moles of Sn in 5.5 g of bronze. The molar mass of Sn is 118.71 g/mol.
Number of moles of Sn = Mass of Sn / Molar mass of Sn
Number of moles of Sn = 5.5 g / 118.71 g/mol
Next, we use the stoichiometric ratio between Sn and H2 to calculate the number of moles of H2 gas evolved. From the balanced equation, we know that 1 mole of Sn reacts to produce 1 mole of H2.
Number of moles of H2 = Number of moles of Sn
Finally, we can use the ideal gas equation to calculate the volume of H2 gas at the given molar volume:
Volume of H2 gas = Number of moles of H2 gas × Molar volume
Volume of H2 gas = Number of moles of H2 gas × 24 L/mol
c) To find the final concentration of Sn^2+ ions, we need to calculate the number of moles of Sn^2+ ions produced and then divide it by the final volume of the solution.
We already know the number of moles of Sn from part b. To find the number of moles of Sn^2+ ions, we use the stoichiometry from the balanced equation. From the equation, it is clear that 1 mole of Sn reacts to produce 1 mole of Sn^2+ ions.
Number of moles of Sn^2+ ions = Number of moles of Sn
Next, we calculate the volume of the solution in liters:
Volume of solution in liters = 200 mL / 1000 mL/L
Finally, we can use the definition of concentration to determine the final concentration of Sn^2+ ions:
Final concentration of Sn^2+ ions = Number of moles of Sn^2+ ions / Volume of solution in liters
a) To write the balanced equation for the reaction between bronze (Cu-Sn) and hydrochloric acid (HCl), we need to know the chemical reactions involved. Bronze is an alloy of copper (Cu) and tin (Sn), so we can represent it as Cu-Sn. When it reacts with hydrochloric acid, the tin (Sn) in the bronze will be converted into tin chloride (SnCl2), while copper (Cu) remains unchanged:
Cu-Sn + 4HCl -> Cu + SnCl2 + 2H2
b) To find the volume of H2 evolved, we need to utilize stoichiometry. The balanced equation tells us that 1 mole of Cu-Sn reacts with 4 moles of HCl to produce 2 moles of H2.
First, we need to determine the number of moles of Cu-Sn in the 20.5g sample. To do this, we divide the mass by the molar mass:
Molar mass of Cu = 63.55 g/mol
Molar mass of Sn = 118.71 g/mol
Molar mass of Cu-Sn = 63.55 + 118.71 = 182.26 g/mol
Number of moles of Cu-Sn = mass / molar mass = 20.5 g / 182.26 g/mol = 0.1124 moles
Since Cu-Sn reacts with HCl in a ratio of 1:4, we know that 0.1124 moles of Cu-Sn will react with 0.1124 * 4 = 0.4496 moles of HCl.
According to the stoichiometry of the reaction, 1 mole of Cu-Sn produces 2 moles of H2. Therefore, 0.4496 moles of HCl will produce 2 * 0.4496 = 0.8992 moles of H2.
Now, to find the volume of H2, we can use the molar volume of the gas, which is 24 L/mol. We can set up a proportion:
(volume of H2) / (0.8992 moles of H2) = (24 L/mol) / (1 mole of H2)
volume of H2 = (0.8992 moles of H2) * (24 L/mol) / (1 mole of H2) = 21.58 L
Therefore, the volume of H2 evolved is 21.58 L.
c) To find the final concentration of Sn2+ ions produced, we need to know the number of moles of SnCl2 produced and the final volume of the solution. From part b), we found that 0.1124 moles of SnCl2 are produced.
Now, we need to convert the volume of the solution from milliliters (mL) to liters (L). Since 1 L is equal to 1000 mL, the volume of the solution is 200 mL / 1000 = 0.2 L.
The concentration of Sn2+ ions can be calculated using the formula:
Concentration (M) = (moles of solute) / (volume of solution in liters)
Concentration of Sn2+ ions = (0.1124 moles) / (0.2 L) = 0.562 M
Therefore, the final concentration of Sn2+ ions produced is 0.562 M.