sodium borohydride ($\text{nabh}_4$) is used industrially in many organic syntheses. one way to prepare it is by reacting sodium hydride with gaseous diborane ($\text{b}_2\text{h}_6$). assuming an 76.1% yield, how many grams of $\text{nabh}_4$ can be prepared by reacting 7.78 g of sodium hydride and 9.47 g of diborane? be sure your answer has the correct number of significant figures.

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{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Chemical Equation Balancing", "Stoichiometry", "Chemical Equation Writing" ], "new_concepts": [ "Percent Yield Calculations", "Limiting Reactant Stoichiometry" ], "current_concepts": [ "Chemical Equation Writing", "Chemical Equation Balancing", "Stoichiometry", "Limiting Reactant Stoichiometry", "Percent Yield Calculations" ] } ## Write and balance the chemical equation Using the *Chemical Equation Writing* and *Chemical Equation Balancing* knowledge points $$ 2\text{NaH}(s) + \text{B}_2\text{H}_6(g) ightarrow 2\text{NaBH}_4(s) $$ ## Calculate the moles of each reactant Using the *Stoichiometry* knowledge point $$ \begin{aligned} M(\text{NaH}) &= 23.997\text{ g/mol}\ n(\text{NaH}) &= \frac{7.78\text{ g}}{23.997\text{ g/mol}} \approx 0.3242\text{ mol}\ M(\text{B}_2\text{H}_6) &= 27.67\text{ g/mol}\ n(\text{B}_2\text{H}_6) &= \frac{9.47\text{ g}}{27.67\text{ g/mol}} \approx 0.3422\text{ mol} \end{aligned} $$ ## Determine the limiting reactant Using the *Stoichiometry* knowledge point $$ \begin{aligned} \text{Required } n(\text{NaH}) &= 2 \times n(\text{B}_2\text{H}_6) = 2 \times 0.3422\text{ mol} = 0.6844\text{ mol}\ 0.3242\text{ mol} &< 0.6844\text{ mol} \implies \text{NaH is the limiting reactant.} \end{aligned} $$ ## Calculate the theoretical yield of sodium borohydride Using the *Stoichiometry* knowledge point $$ \begin{aligned} n(\text{NaBH}_4)_{\text{theoretical}} &= n(\text{NaH}) = 0.3242\text{ mol}\ M(\text{NaBH}_4) &= 37.83\text{ g/mol}\ m(\text{NaBH}_4)_{\text{theoretical}} &= 0.3242\text{ mol} \times 37.83\text{ g/mol} \approx 12.265\text{ g} \end{aligned} $$ ## Calculate the actual yield using percent yield We apply the percent yield formula to find the actual mass of product obtained. $$ \begin{aligned} \text{Percent Yield} &= 76.1\%\ m(\text{NaBH}_4)_{\text{actual}} &= m(\text{NaBH}_4)_{\text{theoretical}} \times 0.761\ m(\text{NaBH}_4)_{\text{actual}} &= 12.265\text{ g} \times 0.761 \approx 9.33\text{ g} \end{aligned} $$ Sodium borohydride ($\text{NaBH}_4$) is used industrially in many organic syntheses. One way to prepare it is by reacting sodium hydride with gaseous diborane ($\text{B}_2\text{H}_6$). Assuming an $76.1\%$ yield, how many grams of $\text{NaBH}_4$ can be prepared by reacting $7.78\text{ g}$ of sodium hydride and $9.47\text{ g}$ of diborane? Be sure your answer has the correct number of significant figures. $9.33$ $\text{g}$ { "subject": "Natural Science", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Natural Science", "Chemistry", "Limiting Reactant Stoichiometry" ] }

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Explanation:

Write and balance the chemical equation

Calculate the moles of each reactant

Determine the limiting reactant

Calculate the theoretical yield of sodium borohydride

Calculate the actual yield using percent yield

Answer: