Question

1. What is the molar mass of CaCO₃? How many moles of CaCO₃are there in 25 g of CaCO₃? How many CaCO₃ molecules are in 25 g CaCO₃?
2. Balance the following equations

C₄H₁₀ +         O₂ à            3 CO₂         +         H₂O

C₂H₆ +         O₂ à            3 CO₂         +       H₂O

How many ml of a 45% HCl solution do you need to prepare 100 ml of 25% HCl solution by diluting?

Answer 1

1) Molar mass of CaCO3 = 100 g/mol

Moles in 25 g of CaCO3 = mass/molar mass

= 25 g/ 100 g/mol

= 0.25 mol

molecules are in 25 g CaCO₃ = moles x Avogadro's number

= 0.25 mol x 6.022 x 10²³ molecules/mol

= 1.505 x 10²³ molecules

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2) General equation for combustion of alkanes is

  CxHy + [ (x)+(y/2) ] O₂ -------------> x CO₂   + (y/2) H₂O

The balanced equations are

C₄H₁₀ + (13/2) O₂ ------------->  4CO₂   + 5 H₂O

C₂H₆ + (7/2) O₂ ------------> 2 CO₂    + 3 H₂O

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3) C1 V1 = C2V2

45% x V1 = 25 % x 100 ml

Then,

V1 = 55.55 ml

Therefore,

55.55 ml of 45% HCl solution need to prepare 100 ml of 25% HCl solution by diluting.