If a scientist weighed 0.68 g of the reactant Na2CO3 for the reaction below, how many moles of Na2CO3 would he have? (Assume Na2CO3 is the limiting reactant and the molar mass of Na2CO3 is 105.99 g/mol) Na2CO3 (s) + 2 HCI (aq) -> 2 NaCl (aq) + H₂O (1) + CO₂ (g) 0.0075 mol 00.0043 mol 072 mol 00.0064 mol
If a scientist weighed 0.68 g of the reactant Na2CO3 for the reaction below, how many moles of Na2CO3 would he have? (Assume Na2CO3 is the limiting reactant and the molar mass of Na2CO3 is 105.99 g/mol) Na2CO3 (s) + 2 HCI (aq) -> 2 NaCl (aq) + H₂O (1) + CO₂ (g) 0.0075 mol 00.0043 mol 072 mol 00.0064 mol
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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Transcribed Image Text:2.
When calculating the yield in a stoichiometry problem, we
generally follow four steps after balancing the chemical equation: 1)
convert all given masses to moles, using conversion factors created
from the molar masses; 2) convert moles of limiting reactant to moles
of product, using conversion factors created from their stoichiometric
coefficients in the chemical equation; 3) calculate the theoretical yield
of the product by converting moles of product to mass of product, using
conversion factors created from its molar mass; 4) calculate the yield of
the product from its experimentally obtained mass and its theoretical
yield.
If a scientist weighed 0.68 g of the reactant Na2CO3 for the reaction
below, how many moles of Na2CO3 would he have? (Assume Na2CO3
is the limiting reactant and the molar mass of Na2CO3 is 105.99 g/mol)
Na₂CO3 (s) + 2 HCI (aq) -> 2 NaCl (aq) + H₂O (1) + CO₂ (g)
0.0075 mol
00.0043 mol
072 mol
00.0064 mol
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