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Chemistry Question on Some basic concepts of chemistry

How many molecules of CO2CO_2 are formed when one milligram of 100%100\% pure CaCO3CaCO_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

6.023×10186.023 \times 10^{18}

Answer

6.023×10186.023 \times 10^{18}

Explanation

Solution

CaCO3100g+2HCl→CaCl2+H2O+CO21 mol\underset{100 g}{CaCO_{3}}+2HCl \to CaCl_2+H_2O+\underset{\text{1 mol}}{CO_{2}}
=6.022×1023=6.022 \times 10^{23} molecules

∵ 100 g CaCO3\because \,100 \,g \, CaCO _{3} gives, molecules of CO2C O _{2}
=6.1022×1023=6.1022 \times 10^{23}
∴ 1×10−3 g CaCO3\therefore \,1 \times 10^{-3} \,g \,CaCO _{3} gives molecules of CO2CO _{2}
=6.022×1023×1×10−3100=\frac{6.022 \times 10^{23} \times 1 \times 10^{-3}}{100}
=6.022×1018=6.022 \times 10^{18}