Solution Found!
Calculate the mass of each product formed when 43.82 g of
Chapter 3, Problem 44P(choose chapter or problem)
Calculate the mass of each product formed when \(43.82 g\) of diborane (\(\mathrm{B}_{2} \mathrm{H}_{6}\)) reacts with excess water:
\(\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{H}_{3} \mathrm{BO}_{3}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g})\) [unbalanced]
Questions & Answers
QUESTION:
Calculate the mass of each product formed when \(43.82 g\) of diborane (\(\mathrm{B}_{2} \mathrm{H}_{6}\)) reacts with excess water:
\(\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{H}_{3} \mathrm{BO}_{3}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g})\) [unbalanced]
ANSWER:
Step 1 of 2
Here we have to calculate the mass of each product formed when 43.82 g of diborane reacts with excess water:
The given unbalanced reaction is,
1st we have to write the balanced chemical equation,
Mass of formed from
:
Given:
Mass of = 48.32 g
Molar mass of = 61.83g/mol
Molar mass of = 27.67 g/mol
Here 2 moles of has formed from 1 mole of
Then the mass of can be calculated as,
Given mass of
= 48.32 g of
= 150.2 g
Thus the mass of is found to be 150.2 g.