Baking soda (NaHCO3) can be made in large quantities by the following reaction: If 10.0 g of NaCl reacts with excesses of the other reactants and 4.2 g of NaHCO3 is isolated, what is the percent yield of the reaction?
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Given: Mass of NaCl = 10.0 g Molar mass of NaCl = 58.44 g/mol Number of moles of NaCl = Mass/Molar mass Number of moles of NaCl = 10.0 g / 58.44 g/mol Number of moles of NaCl = 0.1711 mol Show more…
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Baking soda (NaHCO $_{3}$ ) can be made in large quantities by the following reaction: $\mathrm{NaCl}(a q)+\mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow$.$$\mathrm{NaHCO}_{3}(s)+\mathrm{NH}_{4} \mathrm{Cl}(a q)$$ If $10.0 \mathrm{g}$ of $\mathrm{NaCl}$ reacts with excesses of the other reactants and $4.2 \mathrm{g}$ of $\mathrm{NaHCO}_{3}$ is isolated, what is the percent yield of the reaction?
If 9.29 g of NaCl reacts with excesses of the other reactants and 3.71 g of NaHCO3 is isolated, what is the percent yield of the reaction?
Ravi R.
If baking soda (sodium hydrogen carbonate) is heated strongly, the following reaction occurs: $$2 \mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)$$ Calculate the mass of sodium carbonate that will remain if a $1.52-\mathrm{g}$ sample of sodium hydrogen carbonate is heated.
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