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Svetlanka [38]
3 years ago
10

Consider the chemical reaction: 2 cu(no3)2→ 2 cuo (s) + 4 no2 (g) + o2 (g)when 9.378 g of cu(no3)2 completely decomposed, how

many liters of gas will be produced at 273 k and 1 atm? the molar mass of cu(no3)2 is 187.56 g/mol.a) 0.56
b.1.12c) 2.24
d.2.80e) 3.92
Chemistry
1 answer:
Allushta [10]3 years ago
6 0

2Cu(NO3)2 ---------->    2CuO (s) +4NO2 (g) + O2(g)

9.378g=0.05moles   

no of moles = weight / MW = 9.378/187.56 = 0.05moles

as per the above reaaction 2moles of Cu(NO3)2 can produce 4moles of N2

                                                    0.05moles Cu(NO3)2 can produce (0.05*4)/2 = 0.1moles of N2

and 2moles of Cu(NO3)2 can produce 1moles of O2

       0.05moles Cu(NO3)2 can produce (0.05*1)/2 = 0.025moles of O2

Total moles of gas i.e., N2 and O2 =0.1+0.025 = 0.125moles

From PV = nRT

       V = nRT/ P = 0.125*0.0821*273 = 2.80166Lit option is correct

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zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol

The given reaction equation is as follows.

C + 2H_{2} \rightarrow CH_{4}

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

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Explanation:

This is only the answer if you were asking:

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I think your answer is 2 because the temperature will rise so the particles will move faster
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