Circle the correct answer. Show your work in order to get a full grade. 1. (5 Po

Question

Circle the correct answer. Show your work in order to get a full grade. 1. (5 Points) From the heats of reaction H=-483.6 kJ H=+284.6 kJ - 2 o, (g) 3 02 (g) Calculate the heat of reaction 3 H2O (g) 3 H2 (g) + o, (g) Naco, (s) + H2O (l) 2. (4 Points) 2 NOH (s) + CO2 (g) How many grams of water is produced if 5 g of each reactant is mixed together? 3. (2 Points) How many moles of carbon dioxide are there in 52.06 g of carbon dioxide? A) 0.8452 B) 1.183 C) 6,022 x 102, D) 8648 x 1023 E)3.134 x 103 NO2 and (3 Points) The combustion of ammonia in the presence of excess oxygen yields 4 H,0: 4NH, (g) + 702 (g)4NO2 (g) + 6 H2O(g) The combustion of 28.8 g of ammonia consumes A) 94.9 B) 54.1 C) 108 D) 15.3 E) 28.8 g of oxygen. 5 (3 Points) What mass in grams of hydrogen is produced by the reaction of 4.73 g of magnesium with 1.83 g of water? Mg(s) +2H20 (l) Mg(OH)2 (s) + H2(g) A) 0.102 B) 0.0162 C) 0.0485 D) 0.219 E) 0.204 6. (2 Points) The formula weight of aluminum suifate (Al,(s0,) is A) 342.14 8 123.04 amu.

Explanation / Answer

1) Multiply the first equation with 3 ,
6H2 (g) + 3O2 (g) ---------> 6H2O (g) H = -1450.8 kJ
2O3 (g) --------> 3O2 (g)  H = - 284.6 kJ
Add these above 2 equations,
6H2 (g) + 2O3(g) ------> 6H2O (g)  Hrxn = -1450.8 kJ - 284.6 kJ = -1735.4 kJ
Divide this equation with 2
3H2 (g) + O3 (g) -------> 3 H2O (g)  Hrxn = (-1735.4 kJ) / 2 = -867.7 kJ

2) find the limiting reactant first,
moles of NaOH = 5gm / 39.997 g/mol = 0.1250 mol
moles of CO2 = 5gm / 44.01 g/mol = 0.1136 mol
CO2 is the limiting reactant
According to the reaction stoichiometry,
1 mole of CO2 gives 1 mole of H2O
0.1136 mole of CO2 gives ? moles of H2O
0.1136 mole of H2O (cross multiply to get the answer)
0.1136 mol*18 g/mol = 2.04498 gms of water is produced.

3) moles of CO2 = 52.06 gm / 44.01 gm/mol = 1.1829 mol, option A is the answer

4) moles of NH3 = 28.8 gm / 17.031 g/mol = 1.6910 mol
So combustion of 28.8 gm of NH3 requires 1.6910 mol of O2
1.6910 mol * 32 gm/mol = 54.1 gm of O2 , Option B is the answer.

5) moles of Mg = 4.73 gm / 24.3056 gm/mol = 0.1946 mol
moles of H2O = 1.83 gm / 18 gm/mol = 0.1016 mol
H2O is the limiting reactant ,
According to the reaction stoichiometry,
2 moles of H2O gives 1 mole of H2
0.1016 mol gives ? mol of H2
0.05 mol of H2 is produced (cross multiply to get the answer)
0.05 mol of H2 * 2.016 g/mol = 0.1024 gm of H2  is produced, Option A is the answer

6) [Al2(SO4)3] = 2*27 + 3*32 + 12*16 = 342 amu
Option A is the answer

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