9. Based on the balanced equation below: 8C0 + 17H, ? C8H18 + 8H20 a. How many g
ID: 704644 • Letter: 9
Question
9. Based on the balanced equation below: 8C0 + 17H, ? C8H18 + 8H20 a. How many grams of CsHis would be formed from 2500ml of carbon monoxide, if carbon monoxide has a density of 1.96 g/L (Assuming an excess of Ha)? How many mL of water would be formed from 7500mL of hydrogen gas (Ha), if hydrogen gas has a density of 0.08988 g/L and if water has a density of 0.995g/mL (Assuming an excess of CO)? b. c. If 14.3496g of CaH18 are formed from 14.7L of carbonmonoxi de iunth Hn excess, and assuming 100% year i. what is the experimental density of carbon monoxide, in g/L? what is the percent error of this experimental density? (based upon theoretical density in part "a of this problem) ii.Explanation / Answer
The atomic masses are
C: 12.011 u
H: 1.008 u
O: 15.999 u
The gram molar masses are
CO = (1*12.011 + 1*15.999) g/mol = 28.01 g/mol
H2 = (2*1.008) g/mol = 2.016 g/mol
C8H18 = (8*12.011 + 18*1.008) g/mol = 114.232 g/mol
H2O = (2*1.008 + 1*15.999) g/mol = 18.015 g/mol.
(a) The density of CO = 1.96 g/L; the volume of CO = 2500 mL.
The mass of CO corresponding to 2500 mL = (volume of CO in L)*(density of CO in g/L) = (2500 mL)*(1 L/1000 mL)*(1.96 g/L) = 4.90 g.
Mole(s) of CO corresponding to 4.90 g = (mass of CO)/(molar mass of CO) = (4.90 g)/(28.01 g/mol) = 0.1749 mole.
As per the stoichiometric equation,
8 moles CO = 1 mole C8H18.
Therefore, 0.1749 mole CO = (0.1749 mole CO)*(1 mole C8H18/8 moles CO) = 0.0218625 mole C8H18.
Mass of C8H18 produced = (moles of C8H18)*(molar mass of C8H18) = (0.0218625 mole)*(114.232 g/mol) = 2.4974 g ? 2.50 g (ans).
(b) The density of H2 = 0.08988 g/L; the volume of H2 = 7500 mL.
The mass of H2 corresponding to 7500 mL = (volume of H2 in L)*(density of H2 in g/L) = (7500 mL)*(1 L/1000 mL)*(0.08988 g/L) = 0.6741 g.
Mole(s) of H2 corresponding to 0.6741 g = (mass of H2)/(molar mass of H2) = (0.6741 g)/(2.016 g/mol) = 0.334375 mole.
As per the stoichiometric equation,
17 moles H2 = 8 mole H2O.
Therefore, 0.334375 mole H2 = (0.334375 mole H2)*(8 mole H2O/17 moles H2) = 0.157353 mole H.
Mass of H2O produced = (moles of H2O)*(molar mass of H2O) = (0.157353 mole)*(18.015 g/mol) = 2.8347 g.
Volume of H2O produced in mL = (mass of H2O produced in g)/(density of H2O in g/mL) = (2.8347 g)/(0.995g/mL) = 2.8489 mL ? 2.849 mL (correct to 4 sig. figs, ans).
c) (i) Moles of C8H18 corresponding to 14.3496 g = (mass of C8H18)/(molar mass of C8H18) = (14.3496 g)/(114.232 g/mol) = 0.125618 mole.
As per the stoichiometric equation,
1 mole C8H18 = 8 moles CO.
Therefore, 0.125618 mole C8H18 = (0.125618 mole C8H18)*(8 moles CO/1 mole C8H18) = 1.004944 mole CO.
Mass of CO corresponding to 1.00494 mole CO = (moles of CO)*(molar mass of CO) = (1.004944 mole)*(28.01 g/mol) = 28.148481 g.
Experimental density of CO = (mass of CO)/(volume of CO) = (28.148481 g)/(14.7 L) = 1.9149 g/L ? 1.91 g/L (correct to 3 sig. figs, ans).
ii) The theoretical density of CO is 1.96 g/L and the experimental density is 1.91 g/L. Therefore, percent error in the calculation of density of CO = [(actual density) – (experimental density)]/(actual density)*100 = [(1.96 – 1.91) g/L]/(1.96 g/L)*100 = 0.05/1.96*100 = 2.5510 ? 2.55% (ans).
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