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Question: How many molecules of \[C{{O}_{2}}\] are formed when one milligram of 100% pure \[CaC{{O}_{3}}\,\] i...

How many molecules of CO2C{{O}_{2}} are formed when one milligram of 100% pure CaCO3 CaC{{O}_{3}}\, is treated with excess hydrochloric acid?
(a) 6.023×10236.023\times {{10}^{23}}
(b) 6.023×10216.023\times {{10}^{21}}
(c) 6.023×10206.023\times {{10}^{20}}
(d) 6.023×10196.023\times {{10}^{19}}
(e) 6.023×10186.023\times {{10}^{18}}

Explanation

Solution

Molar mass is the relationship between moles and mass. In other words, the weight in grams of one mole of a substance. We can find the molar mass for any element under the symbol on the periodic table.

Complete step by step solution:
CaCO3+2HCl→CaCl2+H2O+CO2CaC{{O}_{3}}+2HCl\to CaC{{l}_{2}}+{{H}_{2}}O+C{{O}_{2}}
We can see from the equation that 1 mole of calcium carbonate is treated with excess hydrochloric acid to form 1 mole of carbon dioxide. We know that the mass remains conserved in a chemical reaction.
Molar mass of CaCO3 CaC{{O}_{3}}\, can be calculated as:
CaCO3:(1×40 g/mol Ca)+(1×12 g/mol C)+(3×16 g/mol O)=100 g/mol CaCO3CaC{{O}_{3}}:(1\times 40\text{ }g/mol\text{ }Ca)+(1\times 12\text{ }g/mol\text{ }C)+(3\times 16\text{ }g/mol\text{ }O)=100\text{ }g/mol\text{ Ca}C{{O}_{3}}
We will now convert the given mass of CaCO3CaC{{O}_{3}} to moles by dividing by its molar mass.
Moles of CaCO3=1×10−3100=1×10−5 mole CaCO3Moles\,of\,CaC{{O}_{3}}=\dfrac{1\times {{10}^{-3}}}{100}=1\times {{10}^{-5}}\,mole\,CaC{{O}_{3}}
Now, we will multiply CaCO3 CaC{{O}_{3}}\, by the mole ratio between CO2C{{O}_{2}} and CaCO3 CaC{{O}_{3}}\, from the balanced equation. This will give us the mol of CO2C{{O}_{2}} in the reaction
1×10−3 g CaCO3 × 1 mol CaCO3100g CaCo3 × 1 mol CO21 mol CaCO3=1×10−5 mol CO2 1\times {{10}^{-3}}\,g\,CaC{{O}_{3}}\,\times \,\dfrac{1\,mol\,CaC{{O}_{3}}}{100g\,CaC{{o}_{3}}}\,\times \,\dfrac{1\,mol\,C{{O}_{2}}}{1\,mol\,CaC{{O}_{3}}}=1\times {{10}^{-5}}\,mol\,C{{O}_{2}}\,
We know 1 mole of any substance contains Avogadro’s number of molecules or atoms.
Finally, using unitary method 1×10−5 mol CO2 1\times {{10}^{-5}}\,mol\,C{{O}_{2}}\,will give:
Number of molecules of CO2=6.023×1023×1×10−5 mol CO2            =6.023×1018 \begin{aligned} & C{{O}_{2}}=6.023\times {{10}^{23}}\times 1\times {{10}^{-5}}\,mol\,C{{O}_{2}}\, \\\ & \,\,\,\,\,\,\,\,\,\,=6.023\times {{10}^{18}} \\\ \end{aligned}

So, the correct option is (e).

Note: Avogadro’s number is equal to 6.023×10236.023\times {{10}^{23}} units. It is valid for any substance, whether it is an ion, atom or molecule. It is also known as Avogadro’s constant is some literature. Generally speaking, a mole of any substance contains 6.023×10236.023\times {{10}^{23}} particles of that substance.