Answer:
0.466 L (or 0.46648 L if not rounding)
Explanation:
Hey so this problem is just testing your knowledge of stoichiometry with volume analyses and such.
For this first we notice that NaHCO3 is excess so we don't care about it since it doesn't affect the final product.
So we start by finding the moles of acetic acid (Hc2h3o2) because we remember
Molarity=moles/litres
0.833=x/(25/1000)----we divide by thousand to convert ml to litres
x= 0.020825 moles acetic acid x 1mol Co2/1mol acetic acid (this is just using molar ratio to go from moles of acetic acid to moles of co2)
x= 0.020825 mol CO2
Now since it is at STP we know molar volume of a gas at STP (Standard Temperature and Pressure), which is equal to 22.4 L for 1 mole of any ideal gas.
So we just multiply 0.020825mol Co2 x 22.4L/1mol Co2 = 0.46648L or 0.466L
