If $6.3 g$ of $N a H C O_{3}$ are added to $15.0 g$ of $C H_{3} C O O H$ solution, the residue is found to weigh $18.0 g$ . What is the mass of $C O_{2}$ released in the reaction?

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Solution

$N a H C O_{3} + C H_{3} C O O H \to C H_{3} C O O N a + H_{2} O + C O_{2}$
molecular weights ,
$N a H C O_{3} = 84$ , $C H_{3} C O O H = 60$ , $C O_{2} = 44$ , given $6.3$ g of $N a H C O_{3}$ reacts with $15$ g of $C H_{3} C O O H$ .
Number of moles of $N a H C O_{3} =$ given wt. / molecular wt.
$6.3 / 84 = 0.075$ moles
Number of moles of $C H_{3} C O O H = 15 / 60 = 0.25$
it is to be remember that $1$ mole of reactants combines to produce $1$ mole of products.
Thus $0.075$ moles of $N a H C O_{3}$ will reacts with $0.075$ moles of to produce $0.075$ moles of $C O_{2}$
weight of $1$ moles of $C O_{2}$ = $44$ gm
weight of $1$ moles of $C O_{2}$ = $(0.075 \times 44) = 3.3$ gm
so 3.3 gm of $C O_{2}$ will be produced in the reaction

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