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when 86.85 g of diborane (\\text{b}_2\\text{h}_6) reacts with excess wa…

Question

when 86.85 g of diborane (\text{b}_2\text{h}_6) reacts with excess water:

\\\text{b}_2\text{h}_6(g) + \text{h}_2\text{o}(l) \
ightarrow \text{h}_3\text{bo}_3(s) + \text{h}_2(g) \text{ unbalanced}\\

part 1 of 2

calculate the mass of \text{h}_3\text{bo}_3. be sure your answer has the correct number of significant figures.

Explanation:

Balance the chemical equation

Using the Chemical Equation Balancing knowledge point

$$ \text{B}_2\text{H}_6(g) + 6\text{H}_2\text{O}(l) ightarrow 2\text{H}_3\text{BO}_3(s) + 6\text{H}_2(g) $$

Calculate moles of diborane

Using the Stoichiometry knowledge point

$$ LATEXBLOCK0 $$

Determine moles of boric acid

Using the Stoichiometry knowledge point

$$ LATEXBLOCK1 $$

Calculate mass of boric acid

Using the Stoichiometry knowledge point

$$ LATEXBLOCK2 $$

Apply significant figures

Using the Stoichiometry knowledge point

$$ \text{Given mass } 86.85\text{ g has } 4\text{ significant figures.}\\ m(\text{H}_3\text{BO}_3) = 388.1\text{ g} $$

Answer:

Calculate the mass of \(\text{H}_3\text{BO}_3\). Be sure your answer has the correct number of significant figures.

<blank>\(388.1\)</blank> g