B) How many grams of NaN, can be made from 140.05 grams of N,? Balanced Chemical Equation :2 Na + 3 N, → 2 NaN, Step 2: Given 140.05 grams of N,, Use the molar mass to convert into Moles N,. Known Information [Туре] [Type ] Step 3: Use the coefficients of a balanced chemical equation to find the mole ratio [Туре] [Type ] Step 4: Use the mole ratio to convert into Moles of NaN, Mole ratio : (This is conversion factor for step 4) Result from Step 2 [Туре] [Type ] %D 1 Step 5: Convert from moles into final desired unit. [Туре] [Туре ] Result from Step 4 grams NaN3. 1

B) How many grams of NaN, can be made from 140.05 grams of N,? Balanced Chemical Equation :2 Na + 3 N, → 2 NaN, Step 2: Given 140.05 grams of N,, Use the molar mass to convert into Moles N,. Known Information [Туре] [Type ] Step 3: Use the coefficients of a balanced chemical equation to find the mole ratio [Туре] [Type ] Step 4: Use the mole ratio to convert into Moles of NaN, Mole ratio : (This is conversion factor for step 4) Result from Step 2 [Туре] [Type ] %D 1 Step 5: Convert from moles into final desired unit. [Туре] [Туре ] Result from Step 4 grams NaN3. 1

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Solve the stoichiometric problem involving limiting and excess reactants. 1. When 80.0 g of solid copper reacts with 25.0 g of sulfur powder, it will produce a certain amount of solid copper (I) sulfide. Determine: (a) the limiting reactant; (b) the excess reactant; (c) the amount of product formed; and (d) the amount of excess reactant that will remain after the reaction.
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