Question

c₂h₂ + 5o₂ → 4co₂ + 2h₂o
model
atoms
mass: 26g →
does this follow the law of conservation of mass?

Explanation:

Step1: Count atoms in reactants

For \(C\): In \(C_2H_2\), there are \(2\) \(C\) atoms.
For \(H\): In \(C_2H_2\), there are \(2\) \(H\) atoms.
For \(O\): In \(5O_2\), there are \(5\times2 = 10\) \(O\) atoms.

Step2: Count atoms in products

For \(C\): In \(4CO_2\), there are \(4\times1=4\) \(C\) atoms.
For \(H\): In \(2H_2O\), there are \(2\times2 = 4\) \(H\) atoms.
For \(O\): In \(4CO_2\) there are \(4\times2=8\) \(O\) atoms and in \(2H_2O\) there are \(2\times1 = 2\) \(O\) atoms. Total \(O\) atoms \(=8 + 2=10\) atoms.

Step3: Check mass conservation

The balanced equation should be \(2C_2H_2+5O_2
ightarrow4CO_2 + 2H_2O\).
Molar mass of \(C_2H_2=(2\times12)+(2\times1)=26\space g/mol\). For \(2\) moles of \(C_2H_2\), mass \(=2\times26 = 52\space g\).
Molar mass of \(O_2 = 32\space g/mol\). For \(5\) moles of \(O_2\), mass \(=5\times32=160\space g\). Total mass of reactants \(=52 + 160=212\space g\).
Molar mass of \(CO_2=(12)+(2\times16)=44\space g/mol\). For \(4\) moles of \(CO_2\), mass \(=4\times44 = 176\space g\).
Molar mass of \(H_2O=(2\times1)+16 = 18\space g/mol\). For \(2\) moles of \(H_2O\), mass \(=2\times18=36\space g\). Total mass of products \(=176+36 = 212\space g\).

Answer:

Element	# in Reactants	# in Products
\(H\)	\(4\) (from \(2C_2H_2\))	\(4\) (from \(2H_2O\))
\(O\)	\(10\) (from \(5O_2\))	\(10\) (\(8\) from \(4CO_2\) and \(2\) from \(2H_2O\))

Mass of reactants (\(2C_2H_2 + 5O_2\)): \(2\times26+5\times32=212\space g\). Mass of products (\(4CO_2+2H_2O\)): \(4\times44 + 2\times18=212\space g\). So, it follows the Law of Conservation of Mass.