Answer:
197g h2o
Explanation:
Decomposition of Sugar
[tex]C_1_2H_2_2O_1_1[/tex] --->[tex]12C + 11H_2O[/tex]
In order to properly understand the relation between each calculation, lets first calculate each value seperately before solving with dimensional analysis, which is how most teachers would want you to solve it.
Seperate Method
Always start with the given.
You are given the mass of the sugar used for decomposition, which you can use to find out how many moles that is. You want to find the number of moles in order to compare it with the reaction equation, which takes into consideration 1 mole of sugar per reaction.
To find the number of mole(s) of sugar, use the formula
[tex]M(molarmass) = \frac{m(mass)(g)}{mol}[/tex]
mass = 0.750lbs x 453.59g = 340.1925g (3 s.f.)
molar mass = 12(12.011) + 22(1.008) + 11(15.999) = 342.297g/mol (3 dec = 6 s.f.)
mol = m/M
mol = 340.1925 / 342.297 = 0.993852 mol (better to keep longer decimals for more precise answers, to which you can find the sig figs later) (3 s.f.)
Now that you have the number of mol of sugar used, you can find the number of mols of water produced by ratios based on the equation reaction.
[tex]\frac{{1mol_s_u_g_a_r}}{0.993852mol_s_u_g_a_r}\frac{11mol_H_2_O}{xmol_H_2_O}[/tex]
x = 10.932372 mol H2O
Now that you have the number of mol of H2O produced, you find the mass of the H2O just by applying the molar mass formula from earlier.
m = Mn
m = 10.932372*(2(1.008) + 15.999) = 196.94668g H2O
197g H2O (3 s.f.)
DImensional Analysis Method
Dimensional Analysis involves doing all of the steps in one line. So you really have to understand the relation between everything. But it isn't hard. Always start with your given, continue from denominator to numerator until you reach your final numerator, which is the only unit left.
[tex]\frac{340.1925g_s_u_g}{1} *\frac{1mol_s_u_g}{342.297g_s_u_g} *\frac{11mol_h_2_o}{1mol_s_u_g}* \frac{18.01528g_h_2_o}{1mol_h_2_o} = \frac{67415.2946g_h_2_o}{342.297} = 196.9497 (197g_h_2_o)[/tex]