Question

how many grams of Co2 can be produce by thermally decomposing 6.5 moles of CaCO3?​

Answer 1

Answer:

By thermal decomposition of 6.5 moles of CaCO₃, 286 grams of CO₂ are formed

Explanation:

The reaction of decomposition of calcium carbonate is

CaCO₃ -------> CaO + CO₂

It is balanced and it is seen that one mole CaCO₃ produces 1 mole CO₂

So, we can say that 6.5 moles of CaCO₃ will produce 6.5 mole of CO₂.

1 mole CO₂ weighs = (12+2x16) = 12+32=44 g (Where relative atomic masses of C=12gm and O=16gm)

1 mole weighs 44 g

So, 6.5 moles will weigh = (6.5 X 44)gm = 286 gm

Ans) By thermal decomposition of 6.5 moles of CaCO₃, 286 grams of CO₂ are formed

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Answer 2

The amount of $CO_{2}$ produced by thermally decomposing $CaCO_{2}$ is 35.7 g.

Explanation:

• We know the reaction $CaCO_{2a}$ →$CaO_{2} +CO_{2}$↑

• One mole of calcium carbonate will decompose to form one mole of calcium oxide and one mole of carbon dioxide.

• Now, since you did not provide information about the temperature and pressure at which the oxygen gas is being collected, I will assume that you're at STP, Standard Pressure and Temperature.

• Under STP conditions, which are defined as a pressure of  100 kPa and a temperature of $0^{0}$  ,one mole of any ideal gas is known to  occupy 22.7 L -  this is known as the molar volume of a gas at STP.

• This means that you can use the volume of carbon dioxide to figure out how many moles of oxygen were produced by the reaction.

                 $L=\frac{1 mole O_{2} }{22.7 L}*6.5$

                      = 0.2863 moles  .

• Use calcium carbonate's molar mass to find how many grams of calcium carbonate would contain this many moles.

             $=0.2863 moles of CaCO_{2}* \frac{100.1 g}{1 mole CaCO_{2} }\\=28.63$

• Now, you know that your initial sample contained 20 % impurities. This is equivalent to saying that it contained  80 % calcium carbonate.

                 $28.63 g CaCO_{2} =\frac{100 g}{80 g CaCO_{2} }\\=35.7g$

     Therefore, thermally decomposing 6.5 moles produces 35.7 g of $CO_{2}$.