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Posted by on Sunday, September 27, 2009 at 10:22pm.

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 - , Sunday, September 27, 2009 at 10:28pm

    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 - , Sunday, September 27, 2009 at 11:34pm

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

  • Chemistry - , Monday, September 28, 2009 at 12:28am

    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 - , Wednesday, March 21, 2012 at 4:27pm

    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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