How many mL of Carbon Dioxide gas at STP are produced from the decomposition of 4.50g of Ferric Carbonate?
This is How I tried to solve it.....
4.50g Fe12(CO3)3* 1 molFe2(CO3)3/291.72g Fe2(CO3)3* 3 mol CO2/1 mol Fe2(CO3)3* 22.4L CO2/1mol CO2* 1000ml of CO2/1 L of CO2 = 1036.57 which is equal to 1040. mL using 3 sig. figures.
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It is correct.
That looks ok to me.
Your approach to solving the problem is mostly correct. However, there seems to be a mistake in your calculation. Let's break down the calculation step by step:
1. Start with the given mass of Ferric Carbonate (Fe2(CO3)3): 4.50g.
2. Use the molar mass of Ferric Carbonate (Fe2(CO3)3) to convert grams to moles:
4.50g Fe2(CO3)3 * (1 mol Fe2(CO3)3 / 291.72g Fe2(CO3)3) = 0.0154 mol Fe2(CO3)3.
3. Use the balanced chemical equation to determine the mole ratio between Ferric Carbonate (Fe2(CO3)3) and Carbon Dioxide (CO2). From the equation:
Fe2(CO3)3 → 3CO2.
The mole ratio is 1:3.
4. Convert moles of Ferric Carbonate (Fe2(CO3)3) to moles of Carbon Dioxide (CO2) using the mole ratio:
0.0154 mol Fe2(CO3)3 * (3 mol CO2 / 1 mol Fe2(CO3)3) = 0.0462 mol CO2.
5. Use the ideal gas law to convert moles of Carbon Dioxide (CO2) to volume at STP (Standard Temperature and Pressure):
0.0462 mol CO2 * (22.4 L / 1 mol CO2) = 1.034 L CO2.
6. Finally, convert liters to milliliters:
1.034 L CO2 * (1000 mL / 1 L) = 1034 mL CO2.
Therefore, the correct answer is approximately 1034 mL of Carbon Dioxide gas at STP produced from the decomposition of 4.50g of Ferric Carbonate.