Name: Ethan Alor $\newline$ Period: $4$ $\newline$ Limiting Reactant Test $\newline$ Calculate the grams of $\text{NaBr}$ produced when $92.3$ grams of $\text{NBr}_3$ is reacted with $88.8$ grams of $\text{NaOH}$ , according to the following equation (Balance First): $\text{NBr}_3+\text{NaOH}\rightarrow \text{N}_2+\text{NaBr}$ $\newline$ Suppose $143.0\text{g}$ aluminum sulfide reacts with $283.0\text{g}$ of water. What mass of the excess reactant remains? Equation (Balance First): $\text{Al}_2\text{S}_3+\text{H}_2\text{O}\rightarrow \text{Al(OH)}_3 +\text{H}_2\text{S}$

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Q. Name: Ethan Alor

\newline

Period:

4

\newline

Limiting Reactant Test

\newline

Calculate the grams of

\text{NaBr}

produced when

92.3

grams of

\text{NBr}_3

is reacted with

88.8

grams of

\text{NaOH}

, according to the following equation (Balance First):

\text{NBr}_3+\text{NaOH}\rightarrow \text{N}_2+\text{NaBr}

\newline

Suppose

143.0\text{g}

aluminum sulfide reacts with

283.0\text{g}

of water. What mass of the excess reactant remains? Equation (Balance First):

\text{Al}_2\text{S}_3+\text{H}_2\text{O}\rightarrow \text{Al(OH)}_3 +\text{H}_2\text{S}

Balance Chemical Equation: Step $1$ : Balance the chemical equation for the reaction between $\mathrm{NBr}_3$ and $\mathrm{NaOH}$ . $\mathrm{NBr}_3 + 3\,\mathrm{NaOH} \rightarrow 3\,\mathrm{NaBr} + \mathrm{N}_2$
Calculate Molar Mass: Step $2$ : Calculate the molar mass of $\mathrm{NBr}_3$ and $\mathrm{NaOH}$ . $\newline$ Molar mass of $\mathrm{NBr}_3 = 106.9$ (N) $+ 3 \times 79.9$ (Br) $= 346.6 \, \mathrm{g/mol}$ $\newline$ Molar mass of $\mathrm{NaOH} = 23$ (Na) $+ 16$ (O) $+ 1$ (H) $= 40 \, \mathrm{g/mol}$
Determine Limiting Reactant: Step $3$ : Determine the limiting reactant. $\newline$ Moles of $\mathrm{NBr}_3 = \frac{92.3\,\mathrm{g}}{346.6\,\mathrm{g/mol}} = 0.266\,\mathrm{mol}$ $\newline$ Moles of $\mathrm{NaOH} = \frac{88.8\,\mathrm{g}}{40\,\mathrm{g/mol}} = 2.22\,\mathrm{mol}$ $\newline$ Since $3$ moles of $\mathrm{NaOH}$ are needed per mole of $\mathrm{NBr}_3$ , $\mathrm{NaOH}$ is in excess.
Calculate Grams Produced: Step $4$ : Calculate the grams of NaBr produced. $\newline$ Moles of NaBr = $3 \times \text{Moles of } \text{NBr}_3 = 3 \times 0.266 \text{ mol} = 0.798 \text{ mol}$ $\newline$ Molar mass of NaBr = $23 (\text{Na}) + 80 (\text{Br}) = 103 \text{ g/mol}$ $\newline$ Mass of NaBr = $0.798 \text{ mol} \times 103 \text{ g/mol} = 82.194 \text{ g}$
Balance Chemical Equation: Step $5$ : Balance the chemical equation for the reaction between $\text{Al}_2\text{S}_3$ and $\text{H}_2\text{O}$ . $\newline$ $\text{Al}_2\text{S}_3 + 6\text{H}_2\text{O} \rightarrow 2\text{Al(OH)}_3 + 3\text{H}_2\text{S}$
Calculate Molar Mass: Step $6$ : Calculate the molar mass of $Al_2S_3$ and $H_2O$ . $\newline$ Molar mass of $Al_2S_3 = 2\times27$ (Al) + $3\times32$ (S) = $150$ g/mol $\newline$ Molar mass of $H_2O = 2\times1$ (H) + $16$ (O) = $18$ g/mol
Determine Limiting Reactant: Step $7$ : Determine the limiting reactant for the second reaction. $\newline$ Moles of $\text{Al}_2\text{S}_3 = \frac{143.0\,\text{g}}{150\,\text{g/mol}} = 0.953\,\text{mol}$ $\newline$ Moles of $\text{H}_2\text{O} = \frac{283.0\,\text{g}}{18\,\text{g/mol}} = 15.722\,\text{mol}$ $\newline$ Since $6$ moles of $\text{H}_2\text{O}$ are needed per mole of $\text{Al}_2\text{S}_3$ , $\text{H}_2\text{O}$ is in excess.
Calculate Excess Mass: Step $8$ : Calculate the excess mass of $H_2O$ remaining. $\newline$ Moles of $H_2O$ used = $6 \times$ Moles of $Al_2S_3 = 6 \times 0.953 \, \text{mol} = 5.718 \, \text{mol}$ $\newline$ Mass of $H_2O$ used = $5.718 \, \text{mol} \times 18 \, \text{g/mol} = 102.924 \, \text{g}$ $\newline$ Excess mass of $H_2O = 283.0 \, \text{g} - 102.924 \, \text{g} = 180.076 \, \text{g}$