CH4+202-co2 20 A)679 % C) 187 % B)53.6 % D)464% When 305 moles of CH? are mised

Question

CH4+202-co2 20 A)679 % C) 187 % B)53.6 % D)464% When 305 moles of CH? are mised with 500 moles of O the limiting meactant is 41) CH4 202-cO2+2H2O A) H20 B) CH4 C)02 2) what is the correct form of the conversion factor needed to convert the number of moles of H2O to 42)- the number of moles of NH3 produced? Mg3N26) +6H200)-3Mg (OH)2() 2NH3() 2 moles of NH 6 moles of H2O 6 moles of H20 2 moles of NH3 18 g of H2O 17 g of NH3 17 mole of NH3 B) C) D) 18 moles of IH2o 43) In the reaction of silver nitrate with sodium chloride, how many grams of silver chloride will be produced from 100. g of silver nitrate when it is mixed with an excess of sodium chloride? The 43) AgNOs(aa)+NaCl(ag) -AgCI(o)+ NaNosla) A) 107.9 g B) 58.9 8 C) 0.589 g D) 8448 E) 169.9 g 44) combustion of excess octane with 100.0 g of oxygen assuming complete combustion. The molar mass of octane is 114.33 g/mole. The molar mass of oxygen is 31.9988 g/mole. 14) Using the following equation for the combustion of octane, calculate the heat associated with the Affnn -11018 2 C8H18 + 25 O2-16 CO2 + 18 H2O A) -11018 kJ B)-17220 k) C) -1337 kJ D) -4304 k E)-2152 kJ

Explanation / Answer

41.CH4   +   2O2   ---->   CO2   +   2H2O

In the above reaction equation,

1 mole of CH4 reacts with 2 moles of O2.

Or, 3.05 mole of CH4 reacts with 6.10 moles of O2.

But moles of O2 =5.03 mol

So, O2 is the limiting reagent.

Answer is option (C) O2.

42

Mg3N2(s)   +   6H2O(l)   ----->    3Mg(OH)2(s)   +   2NH3(g)

In the above reaction equation,

6 mole of H2O produces 2 moles of NH3.

Answer is option (C).

43.

AgNO3(aq)   +   NaCl(aq)   ---->   AgCl(s)   +   NaNO3(aq)

In the above reaction equation,

1 mole of AgNO3 produces 1 moles of AgCl.

100 g of AgNO3
Molar mass of AgNO3 = 170 g/mol
So, 170 g of AgNO3 = 1 mol
Or, 1 g of AgNO3 = (1/170) mol
Or, 100 g of AgNO3 = (100/170) mol = 0.59 mol

So, moles of AgCl produced = 0.59 mol
Molar mass of AgCl = 143.32 g/mol
So, 1 mole of AgCl = 143.32 g
or, 0.59 mole of AgCl = 0.59 x 143.32 g = 84.6

Answer is option (D). 84.4 g

44.

2C8H18   +   25O2   ----> 16CO2   +   18   H2O    dHrxn = -11018 kJ

In the above reaction,

25 moles of O2 produces = -11018 kJ

Or, 1 moles of O2 produces = (-11018 / 25) kJ

Or, 31.9988 g of O2 produces = (-11018 / 25) kJ

Or, 1 g of O2 produces = -11018 / (25 x 31.9988) kJ

Or, 100 g of O2 produces = (-11018 x 100) / (25 x 31.9988) kJ = - 1377 kJ

Answer is option (C)

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