Chapter 8.4, Problem 8.4.4SR

Interpretation Introduction

Interpretation:

The mass of water produced in the given reaction should be calculated if the $\text{50g}$ of ${\text{CH}}_{\text{3}}\text{OH}$ undergoes reaction with excess oxygen and the percent yield of the reaction is $\text{53}\text{.2\%}$.

Concept introduction:

• Mole ratios between the reactants and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.
• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Number of moles of a substance, from its given mass is,

  $\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

• Number of grams of a substance from its number of moles is,

  $\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Percent yield of reaction is the ratio of mass of actual yield to the mass of theoretical yield and multiplied with hundred.