Carbon (C) reacts with chlorine (Cl_2) to form tetrachloromethane, CCl_4. A quan
ID: 993657 • Letter: C
Question
Carbon (C) reacts with chlorine (Cl_2) to form tetrachloromethane, CCl_4. A quantity of 10.0 g of carbon is contacted with a quantity of 100.0 g of chlorine. The amount of CCl_4 recovered at the end of the reaction is found to be 89.5 g. The percent yield of CCl_4 is: 34.9% 58.2% 69.9% 82.5% 89.5% None of the above; the correct yield is g The following molecular reaction takes place when a solution of calcium nitrate is contacted with a solution of sodium carbonate: Ca(N0_3)_2(aq) + Na_2 CO_3(aq) rightarrow CaCO_3(s) + 2 NaNO_3(aq) The corresponding net ionic equation is: The following molecular reaction takes place when a solution of sodium hydroxide is contacted with a solution of ammonium chloride: NaOH(aq) + NH_4 Cl(aq) rightarrow NH_3(g) + H_2 O(l) + NaCl(aq) The corresponding net ionic equation is:Explanation / Answer
(7)
C + 2Cl2 -----> CCl4
mass of carbon, C = 10.0g
mass of chlorine, Cl2 = 100.0 g
actual yield of CCl4 given = 89.5 g
moles of C = 10.0 g x 1mol/12g = 0.83 mol
moles of Cl2 = 100.0 g x 1mol/70.92g = 1.41 mol
from balanced equation, 1 mol of C reacts with 2 moles of Cl2, So, moles of Cl2 required to react with 0.83 moles of C could be calculated as...
0.83 mol C x 2 mol Cl2/1 mol C = 1.66 mol Cl2
we need 1.66 moles of Cl2 but only 1.41 mol of it are available, means it is the limiting reactant and product theoretical tield could be calculated by it.
1.41 mol Cl2 x 1 mol CCl4/2mol Cl2 = 0.705 mol CCl4
now moles could be converted into grams by multiply the moles by molar mass.
0.705 mol CCl4 x 153.84 g/1mol = 108.46 g CCl4 (This is theoretical yield)
percent yield = (actual yield/theoretical yield) x 100
percent yield = (89.5/108.46) x 100 = 82.5% (Answer)
So, the correct choice is (D).
(8)
given molecular equation is...
Ca(NO3)2(aq) + Na2CO3(aq) ------> CaCO3(s) + 2NaNO3(aq)
for ionic equations all (aq) species are broken into their ions.
Ca+2(aq) + 2(NO3)-(aq) + 2Na+(aq) + (CO3)-2(aq) -----> CaCO3(s) + 2Na+(aq) + 2(NO3)-(aq)
to get the net ionic we cancel out the spectator (common) ions.
Ca+2(aq) + (CO3)-2(aq) -----> CaCO3(s) (This is the net ionic equation)
(9)
given molecular equation is....
NaOH(aq) + NH4Cl(aq) -----> NH3(g) + H2O(l) + NaCl(aq)
Ionic equtaion.....
Na+(aq) + OH-(aq) + (NH4)+(aq) + Cl-(aq) ------> NH3(g) + H2O(l) + Na+(aq) + Cl-(aq)
net ionic equation....
OH-(aq) + (NH4)+(aq) ------> NH3(g) + H2O(l)