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Calculate the mass of each product formed when 35.56 g of diborane (B2H6) reacts with excess water:

B2H6(g)+H2O(l)=H3BO3(s)+H2(g) [unbalanced]
Mass of H3BO3:_____g
Mass of H2:_____g

User Munichong
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1 Answer

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Final answer:

To calculate the mass of each product formed when 35.56 g of diborane (B2H6) reacts with excess water, we can use the balanced chemical equation to determine the mole ratios and convert grams to moles using molar mass.

Step-by-step explanation:

To calculate the mass of each product, we need to balance the chemical equation first:

B2H6(g) + H2O(l) → 2 H3BO3(s) + 6 H2(g)

From the balanced equation, we can see that 1 mole of B2H6 reacts to form 2 moles of H3BO3 and 6 moles of H2.

Given that we have 35.56 g of B2H6,

Mass of H3BO3 = (35.56 g B2H6)(2 mol H3BO3 / 1 mol B2H6)(61.82 g H3BO3 / 1 mol H3BO3) = 47.92 g H3BO3

Mass of H2 = (35.56 g B2H6)(6 mol H2 / 1 mol B2H6)(2.02 g H2 / 1 mol H2) = 429.67 g H2

User John Forbes
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