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Calculate the grams of
NaBr produced when 92.3 grams of
NBr_(3) is reacted with 88.8 grams of
NaOH, according to the following equation (Balance First):

NBr_(3)+NaOHrarrN_(2)+NaBr

$1$ . Calculate the grams of $\mathrm{NaBr}$ produced when $92$ . $3$ grams of $\mathrm{NBr}_{3}$ is reacted with $88$ . $8$ grams of $\mathrm{NaOH}$ , according to the following equation (Balance First): $\newline$ $\mathrm{NBr}_{3}+\mathrm{NaOH} \rightarrow \mathrm{N}_{2}+\mathrm{NaBr}$

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Partial sums of arithmetic series

Full solution

Q.

1

. Calculate the grams of

\mathrm{NaBr}

produced when

92

.

3

grams of

\mathrm{NBr}_{3}

is reacted with

88

.

8

grams of

\mathrm{NaOH}

, according to the following equation (Balance First):

\newline

\mathrm{NBr}_{3}+\mathrm{NaOH} \rightarrow \mathrm{N}_{2}+\mathrm{NaBr}

Balance Chemical Equation: Balance the chemical equation. $\mathrm{NBr}_3 + 3\,\mathrm{NaOH} \rightarrow 3\,\mathrm{NaBr} + \mathrm{N}_2$
Calculate Molar Mass: Calculate the molar mass of $\mathrm{NBr}_3$ and $\mathrm{NaOH}$ . $\newline$ Molar mass of $\mathrm{NBr}_3 = 1(14.01) + 3(79.90) = 253.71 \, \mathrm{g/mol}$ $\newline$ Molar mass of $\mathrm{NaOH} = 22.99 + 15.999 + 1.008 = 39.997 \, \mathrm{g/mol}$
Determine Moles Used: Determine the moles of $NBr_3$ and $NaOH$ used. $\newline$ Moles of $NBr_3 = \frac{92.3\,\text{g}}{253.71\,\text{g/mol}} = 0.3637\,\text{mol}$ $\newline$ Moles of $NaOH = \frac{88.8\,\text{g}}{39.997\,\text{g/mol}} = 2.2205\,\text{mol}$
Identify Limiting Reactant: Identify the limiting reactant. $\newline$ From the balanced equation, $1$ mol of $\mathrm{NBr}_3$ reacts with $3$ mol of $ext{NaOH}$ . $\newline$ Required $ext{NaOH}$ for $0.3637$ mol of $\mathrm{NBr}_3 = 0.3637 \times 3 = 1.0911$ mol $\newline$ Since $2.2205$ mol of $ext{NaOH}$ is available, which is more than $1.0911$ mol, $\mathrm{NBr}_3$ is the limiting reactant.
Calculate NaBr Produced: Calculate the amount of NaBr produced. $\newline$ From the balanced equation, $1$ mol of $\mathrm{NBr}_3$ produces $3$ mol of $ext{NaBr}$ . $\newline$ Moles of $ext{NaBr}$ produced = $0.3637$ mol $ext{NBr}_3 \times 3 = 1.0911$ mol $ext{NaBr}$
Calculate Mass Produced: Calculate the mass of NaBr produced. $\newline$ Molar mass of NaBr = $22.99 + 79.90 = 102.89 \, \text{g/mol}$ $\newline$ Mass of NaBr = $1.0911 \, \text{mol} \times 102.89 \, \text{g/mol} = 112.23 \, \text{g}$

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