Could someone please help me with the following problem? Thanks in advance!

thread moved. Please post homework in the "homework help" forum

NOTE:- i have taken arbitrary atomic mass to make it easier to explain.learn about limiting reagent and excess reagent once you finish reading this. hope this helps you.

 

let atomic mass of boron be = 2g/mol

hydrogen = 1g/mol

oxygen = 3g/mol

therefore, molecular mass of B5H9= 19g // i will takr it to be 20, for easy explanation

O2 = 6g

H2O = 5g

now,

no. of moles of B5H9 taken = 126/20=6.3

no. of moles of O2 taken = 192/6= 32

according to he rxn.

2 moles of B5H9 will require 12 moles of O2

so 6.3 " " " " " (6*6.3)" " " =37.8 =38(approx)

but we have only 32 moles of O2. hence O2 is the limiting reagent.32 moles of O2 will completly react.

gong back to the rxn.

12 moles of O2 will give 9 moles of water

32 " " " " " 24 " " "

therefore amuont of water produced= 24*5=120 g.

OK first check to see how many moles of each reactant you have. Check with the formula to see which one is the limiting reactant. Then use the stoichiometric coefficients to determine the maximum number of moles of water it's possible to make, finally, convert this to mass.

 

Any problems, just ask. Next time, try to be a bit more specific about what the trouble is, exactly.

 

popcorn, please don't give out answers in the homework help section. read the rules

Could someone please comment on the following solution:

 

[MATH]126 \ \text{g B}_5\text{H}_9 \cdot \frac{1 \ \text{mol} \ \text{B}_5\text{H}_9}{63.14 \ \text{g B}_5\text{H}_9}=2.00 \ \text{mol B}_5\text{H}_9[/MATH]

 

[MATH]192\ \text{g O}_2 \cdot \frac{1 \ \text{mol} \ \text{O}_2}{31.99 \ \text{g O}_2}=6.00 \ \text{mol O}_2 \Rightarrow \times \frac{2 \ \text{mol B}_5\text{H}_9}{12 \ \text{mol O}_2}=0.5 \ \text{mol B}_5\text{H}_9[/MATH]

 

Thus, [MATH]\text{O}_2[/MATH] is the limiting reactant.

 

[MATH]192 \ \text{g O}_2 \cdot \frac{1 \ \text{mol O}_2}{31.99 \ \text{g O}_2} \cdot \frac{9 \ \text{mol H}_2\text{O}}{1 \ \text{mol O}_2} \cdot \frac{18.01 \ \text{g H}_2\text{O}}{1 \ \text{mol H}_2\text{O}}=973 \ \text{g H}_2\text{O}[/MATH]