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For each of the reactions, calculate how many grams of the product form when 2.4 g of
Chapter 8, Problem 33P(choose chapter or problem)
For each of the reactions, calculate how many grams of the product form when 2.4 g of the reactant in color completely reacts. Assume there is more than enough of the other reactant.
(a) \(2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{NaCl}(\mathrm{s})\)
(b) \(\mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(\mathrm{~s})\)
(c) \(2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{MgO}(\mathrm{s})\)
(d) \(\mathrm{Na}_{2} \mathrm{O}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{NaOH}(a q)\)
Questions & Answers
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QUESTION:
For each of the reactions, calculate how many grams of the product form when 2.4 g of the reactant in color completely reacts. Assume there is more than enough of the other reactant.
(a) \(2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{NaCl}(\mathrm{s})\)
(b) \(\mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(\mathrm{~s})\)
(c) \(2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{MgO}(\mathrm{s})\)
(d) \(\mathrm{Na}_{2} \mathrm{O}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{NaOH}(a q)\)
ANSWER:
Step 1 of 4
For each of the reactions. we have to calculate the amount of product in grams form when 2.4 g of the reactant in color completely reacts.
(a)
Chemical reaction:
\(2\mathrm{\ Nas}+\mathrm{Cl}_2(\mathrm{g})\ \rightarrow\ 2\ \mathrm{NaCl}(\mathrm{s})\)
Given: \(2.4\ \mathrm{g\ Cl}_2\)
Find: \(\mathrm{g\ NaCl}\)
\(\mathrm{M}(\mathrm{NaCl})=58.44 \frac{\mathrm{g}}{\mathrm{mol}} \text { and } \mathrm{M}\left(\mathrm{Cl}_{2}\right)=70.9 \frac{\mathrm{g}}{\mathrm{mol}}\)
Solution map: \(\mathrm{g\ Cl}_2\ \rightarrow\mathrm{\ mol\ Cl}_2\ \rightarrow\ \mathrm{mol\ NaCl}\ \rightarrow\mathrm{\ g\ NaCl}\)
\(\frac{1\mathrm{\ mol\ Cl}_2}{70.9\mathrm{\ g\ Cl}_2}\ \rightarrow\ \frac{2\mathrm{\ mol\ NaCl}}{1\ \mathrm{mol\ Cl}_2}\ \rightarrow\ \frac{58.44\mathrm{\ g\ NaCl}}{1\mathrm{\ mol\ NaCl}}\)
\(2.4\ \mathrm{g\ Cl}_2\times\frac{1\mathrm{\ mol\ Cl}_2}{70.9\mathrm{\ g\ Cl}_2}\times\frac{2\mathrm{\ mol\ NaCl}}{1\mathrm{\ mol\ Cl}_2}\times\frac{58.44\ \mathrm{g\ NaCl}}{1\mathrm{\ mol\ NaCl}}=3.96\mathrm{\ g\ NaCl}\)
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