Chapter 16, Problem 16.126QP

Interpretation Introduction

Interpretation:

The volume of ${\text{CO}}_{\text{2}}$(in litres) generated from 0.350 g of ${\text{NaHCO}}_{\text{3}}$ and excess of gastric juice at 1.00 atm and ${37.0}^{\circ }\text{C}$ has to be calculated.

Concept Information:

Mole can be calculated by mass of the substance or solution divided by molar mass of the substance or solution.

$\text{n}=\frac{\text{m}}{\text{M}}$

Where,

M is Molar mass of the substance

m is mass of the substance

Ideal gas law gives the relation between mole, pressure, volume and temperature.

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{where:}\\ \text{P is pressure}\\ \text{V is volume}\\ \text{T is temperature}\\ \text{n is the amount of substance}\\ \text{R is gas constant (8}{\text{.314 J mol}}^{\text{-1}}{\text{ K}}^{\text{-1}}\text{ or 0}{\text{.08206 L atm K}}^{\text{-1}}{\text{ mol}}^{\text{-1}})\end{array}$

To Calculate: The volume of ${\text{CO}}_{\text{2}}$(in litres) generated from 0.350 g of ${\text{NaHCO}}_{\text{3}}$ and excess of gastric juice at 1.00 atm and ${37.0}^{\circ }\text{C}$