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aleksandr82 [10.1K]
2 years ago
11

An industrial manufacturer wants to convert 175 kg of methane into HCN. Calculate the masses of ammonia and molecular oxygen req

uired for this synthesis.
Chemistry
1 answer:
Tema [17]2 years ago
4 0
Given:

175 kilograms of Methane (CH4) to be synthesized into Hydrogen Cyanide (HCN)

The balanced chemical equation is shown below:

2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To calculate for the masses of ammonia and oxygen needed, our basis will be 175 kg CH4.

Molar mass:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol

mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol 
mass of NH3 = 185.94 kg NH3 needed

mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
mass of O2 = 525 kg

mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
mass of O = 131.25 kg O 
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Answer:

percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

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1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

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since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

Theoretical yield of NaCl = 585/84

Theoretical yield of NaCl  = 6.9642857143 g

percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%

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The graph is not given in the question, so, the required graph is attached below:

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Which of the following can be best explained by temporary dipole moments?
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Answer:

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