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How many grams of calcium carbonate will I need to form 3.45 liters of carbon dioxide?CaCO3=CO2+CaO(Performed at STP)

User Chama
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Step-by-step explanation:

CaCO₃ ----> CO₂ (g) + CaO

We have to find the grams of calcium carbonate needed to from 3.45 L of CO₂.

First we can find the number of moles of CO₂ that we have to produce. One mol of any gas at STP occupies 22.4 L. We can use this relationship to find the number of moles present in 3.45 L of CO₂.

1 mol of CO₂ = 22.4 L

moles of CO₂ = 3.45 L * 1 mol/(22.4 L)

moles of CO₂ = 0.154 mol

Once we found the number of moles of CO₂, we can find the number of moles of CaCO₃ that are necessary to produce them. According to the coefficients of the reaction, 1 mol of CO₂ will be produced by 1 mol of CaCO₃.

1 mol of CaCO₃ = 1 mol of CO₂

moles of CaCO₃ = 0.154 moles of CO₂ * 1 mol of CaCO₃/(1 mol of CO₂)

moles of CaCO₃ = 0.154 moles

Finally we can convert these moles into grams using the molar mass of CaCO₃.

molar mass of Ca = 40.08 g/mol

molar mass of C = 12.01 g/mol

molar mass of O = 16.00 g/mol

molar mass of CaCO₃ = 1 * 40.08 g/mol + 1 * 12.01 g/mol + 3 * 16.00 g/mol

molar mass of CaCO₃ = 100.09 g/mol

mass of CaCO₃ = 0.154 moles * 100.09 g/mol

mass of CaCO₃ = 15.4 g

Answer: 15.4 g of CaCO₃ will be needed.

User Michal Kuklis
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