Answer: 1.98 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}[/tex]
[tex]\text{Moles of} CO_2=\frac{5.0L}{22.4L}=0.22moles[/tex]
The balanced given equation is:
[tex]C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)[/tex]
According to stoichiometry :
4 moles of [tex]CO_2[/tex] will produce = 2 moles of [tex]H_2O[/tex]
Thus 0.22 moles of [tex]CO_2[/tex] will produce=[tex]\frac{2}{4}\times 0.22=0.11moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]H_2O=moles\times {\text {Molar mass}}=0.11moles\times 18g/mol=1.98g[/tex]
Thus 1.98 g of water is produced along with 5.0 L of [tex]CO_2[/tex] at STP
