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AlbrechtRLA
Question #2 on Chapter 4 AP Multiple Choice Problems
Sep 28 2009, 7:43 PM EDT | Post edited: Sep 28 2009, 7:43 PM EDT
Can any one explain how to get the answer to question 2 in the ch. 4 multiple choice ws? Do you find this valuable?    

Beata17
1. RE: Question #2 on Chapter 4 AP Multiple Choice Problems
Sep 28 2009, 9:42 PM EDT | Post edited: Sep 28 2009, 9:42 PM EDT
well...
molar mass of carbon: 12.01 g/mol
molar mass of hydrogen: 1.00797 g/mol

so...
balance equation with regards to n=1

2 CN(2) + 3 O(2) = 2 H(2)0 + 2 CO(2)

then add n in front of it because it will change by that much:

2n C(n)N(2n) + 3n O(2) = 2n H(2)0 + 2n CO(2)

so set up a proportion

28n g of C(n)H(2n) / 0.561 g of C(n)H(2n)= 36n g of water/x

the n's cancel out.

solve for x
x=0.72 g of H(2)0

change into moles
=0.0400 moles of H(2)O



does that help? can someone else double check it to make sure i did it right. thanks :) 1  out of 1 found this valuable. Do you?    

Jerrycom42
2. RE: Question #2 on Chapter 4 AP Multiple Choice Problems
Sep 28 2009, 9:50 PM EDT | Post edited: Sep 28 2009, 9:50 PM EDT
First, I balanced the equation and had 2CnH2n + 302 yields CO2 + 2H2O. (the n's and numbers after elements are subscripts) Then i solved for the molar mass of CH2. Since we have n as a subscript for both C and H, it is possible to look at is as n(CH2). If we use different n's and balance the equations and solve for # of moles of H20 formed starting with .561 g CnH2n we get around .04 moles everytime. 1  out of 1 found this valuable. Do you?