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Calculate the mass of each product formed when 43.82g of diborane (B2H6) reacts with excess water:
B2H6(g) + H2O(l)--> H3BO3(s) + H2(g)

  • Chemistry -

    Convert 43.82 g diborane to moles. #moles = grams/molar mass.

    Using the coefficients in the balanced equation, convert moles diborane to moles H3BO3. Do the same for moles H2.

    Convert moles H3BO3 to grams. Convert moles H2 to grams. #grams = moles x molar mass.

  • Chemistry -

    Would the balanced equation for this be:
    B2H6(g)+ 3H2O(l)--> 2H3BO3(s)+ 3H2O(g)

  • Chemistry -

    B2H6(g)+ 3H2O(l)--> 2H3BO3(s)+ 3H2O(g)
    Did you make a typo. There is H2O on both sides. The original equation was
    B2H6 + 3H2O ==>H3BO3 + H2

    B2H6 + 3H2O ==> 2H3BO3 + 3H2 will work.
    Check my work. It's getting late.

  • Chemistry -

    the balanced chemical equation is
    B2H6 + 6H2O -> 2H3BO3 + 6H2

    B2H6 + 3H2O ==> 2H3BO3 + 3H2
    gives you 3O on one side and 6O on the product side.

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