Context:
175 kilograms of methane (CH4) is to be converted into hydrogen cyanide (HCN)
The equation that balances this reaction is listed here:
2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To find the quantities of ammonia and oxygen needed, we will use 175 kg of CH4 as our reference.
Molar masses are as follows:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol
For ammonia: mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol
This results in 185.94 kg of NH3 required
For oxygen: mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
So the mass of O2 needed equals 525 kg
To derive the mass of oxygen: mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
This gives a mass of O equal to 131.25 kg O
Answer:
There's a lot to address, so I'm uncertain if I can tackle this; it feels overwhelming. Perhaps you could simplify it for me as it's quite extensive.
Explanation:
- I wish I had the answers.
- It's too complex.
- This chemistry isn't familiar to me.
- If you have another chemistry-related question, feel free to ask.
- This is just too difficult.
The calculation of moles of chromium (III) nitrate produced is done as follows. First, you write the reaction equation: 3 Pb(NO3)2 + 2 Cr = 2 Cr(NO3)3 + 3 Pb. Then, by using the mole ratio from Pb(NO3)2 to Cr(NO3)3, which is 3 to 2, you can find the moles of Cr(NO3)3. Thus, for 0.85 moles of lead (IV) nitrate, it equates to 0.85 x 2 / 3 = 0.57 moles.
Science is ever-evolving with continuous discoveries; thus, the statement "it is stable and does not require modification due to new findings" is incorrect.
The answer is that the diamond's volume measures 0.063 ml.
With a density of d(diamond) = 3.51 g/ml, and a mass of m(diamond) = 1.11 carat, with 1 carat being equivalent to 0.2 grams, we convert m(diamond) to grams: m(diamond) = 1.11 carat·0.2 gram/carat, which gives m(diamond) = 0.222 g.
To find the volume: V(diamond) = m(diamond) ÷ d(diamond), which results in V(diamond) = 0.222 g ÷ 3.51 g/ml = 0.063 ml.
