CaCO3 + H2
SO4 ------> CaSO4 + CO2 + H2
O
200 g of calcium carbonate are added to 4 L of water.
a. What is the molarity of the calcium carbonate solution?
b. How many moles of hydrogen sulfate would be needed to
react with a 25 mL sample of the calcium carbonate solution?
c. If you added 10 mL of hydrogen sulfate solution, what is the
molarity of the hydrogen sulfate solution?
d. If you are at STP, what volume of carbon dioxide would be
produced?
NaHCO3 + HCl ----> NaCl + CO2 + H2
O
150 g of sodium bicarbonate are added to 2 L of water.
a. What is the molarity of the sodium bicarbonate solution?
b. How many moles of hydrogen chloride would be needed to
react with a 35 mL sample of the sodium bicarbonate solution?
c. If you added 15 mL of hydrogen chloride solution, what is the
molarity of the hydrogen chloride solution?
d. If you are at STP, what volume of carbon dioxide would be
produced?
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a. To find the molarity of the calcium carbonate solution, we first need to calculate the number of moles of calcium carbonate.
The molar mass of calcium carbonate (CaCO3) can be found by summing the atomic masses of its individual elements:
M(Ca) = 40.08 g/mol
M(C) = 12.01 g/mol
M(O) = 16.00 g/mol
M(CaCO3) = M(Ca) + M(C) + 3 * M(O)
= 40.08 + 12.01 + 3 * 16.00
= 100.09 g/mol
Next, we need to determine the number of moles of calcium carbonate in the given 200 g of the compound:
moles of CaCO3 = mass of CaCO3 / Molar mass of CaCO3
= 200 g / 100.09 g/mol
= 1.999 moles (approximately)
Since the solution is prepared in 4 L of water, the molarity (M) can be calculated using the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
Molarity of calcium carbonate solution = 1.999 moles / 4 L
= 0.4997 M (approximately)
Therefore, the molarity of the calcium carbonate solution is approximately 0.4997 M.
b. To determine the number of moles of hydrogen sulfate required to react with a 25 mL sample of the calcium carbonate solution, we need to use the balanced chemical equation provided.
According to the balanced equation:
1 mole of CaCO3 reacts with 1 mole of H2SO4
Since the molarity of the calcium carbonate solution is 0.4997 M, the number of moles of CaCO3 in a 25 mL sample can be calculated as follows:
moles of CaCO3 = Molarity * volume (in liters)
= 0.4997 M * (25 mL / 1000 mL/L)
= 0.0125 moles
Therefore, 0.0125 moles of hydrogen sulfate would be needed to react with the 25 mL sample of the calcium carbonate solution.
c. To find the molarity of the hydrogen sulfate solution after adding 10 mL of the solution, we first need to know the initial molarity of the hydrogen sulfate solution.
Unfortunately, the given information does not provide the initial molarity of the hydrogen sulfate solution. Without the initial molarity, we cannot accurately calculate the final molarity after adding the 10 mL of the solution. Additional information is needed to solve this part of the question.
d. To determine the volume of carbon dioxide produced at STP, the ideal gas law can be used. At STP conditions (Standard Temperature and Pressure), the pressure is 1 atm and the temperature is 273 K.
According to the balanced chemical equation:
1 mole of CaCO3 reacts to produce 1 mole of CO2
Since we already have calculated the number of moles of CaCO3 in the solution (1.999 moles), we know that the same number of moles of CO2 will be produced.
Using the ideal gas law: PV = nRT
where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is temperature in Kelvin.
Rearranging the formula to solve for volume:
V = nRT / P
= (1.999 moles) * (0.0821 L·atm/(mol·K)) * (273 K) / (1 atm)
≈ 44.58 L
Therefore, at STP conditions, approximately 44.58 L of carbon dioxide would be produced.