Asked by TeeJay
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)
B2H6(g) + H2O(l)--> H3BO3(s) + H2(g)
Answers
Answered by
DrBob222
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.
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.
Answered by
Jamie
Would the balanced equation for this be:
B2H6(g)+ 3H2O(l)--> 2H3BO3(s)+ 3H2O(g)
B2H6(g)+ 3H2O(l)--> 2H3BO3(s)+ 3H2O(g)
Answered by
DrBob222
B2H6(g)+ 3H2O(l)--> 2H3BO3(s)+ 3H2O(g)
<b>Did you make a typo. There is H2O on both sides. The original equation was
B2H6 + 3H2O ==>H3BO3 + H2
</b>
<i>B2H6 + 3H2O ==> 2H3BO3 + 3H2 will work.
Check my work. It's getting late.</i>
<b>Did you make a typo. There is H2O on both sides. The original equation was
B2H6 + 3H2O ==>H3BO3 + H2
</b>
<i>B2H6 + 3H2O ==> 2H3BO3 + 3H2 will work.
Check my work. It's getting late.</i>
Answered by
Stubbulus
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.
B2H6 + 6H2O -> 2H3BO3 + 6H2
B2H6 + 3H2O ==> 2H3BO3 + 3H2
gives you 3O on one side and 6O on the product side.
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