The mass of sodium chloride at the two parts are mathematically given as
- m=10,688.18g
- mass of Nacl(m)=39.15g
<h3>What is the mass of sodium chloride that can react with the same volume of fluorine gas at STP?</h3>
Generally, the equation for ideal gas is mathematically given as
PV=nRT
Where the chemical equation is
F2 + 2NaCl → Cl2 + 2NaF
Therefore
1.50x15=m/M *(1.50*0.0821)
1-50 x 15=m/58.5 *(1.50*0.0821)
m=10,688.18g
Part 2
PV=m'/MRT
1*15=m'/58.5*0.0821*273
m'=39.15g
mass of Nacl(m)=m'=39.15g
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Answer:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
Answer:
Explanation:
The air 9% mole% methane have an average molecular weight of:
9%×16,04g/mol + 91%×29g/mol = 27,8g/mol
And a flow of 700000g/h÷27,8g/mol = 25180 mol/h
In the reactor where methane solution and air are mixed:
In = Out
Air balance:
91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)
Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>
<em></em>
The air in the product gas is
95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>
43058 mol air×29g/mol <em>1249 kg air</em>
Percent of oxygen is: =<em>0,231 kg O₂/ kg air</em>
<em></em>
I hope it helps!
Nonpolar covalent bonds are chemical bonds where two atoms share a pair of electrons with each other and the electronegativities of the two atoms are equal. An example is methane. It has four carbon-hydrogen single covalent bonds. These bonds are nonpolar because the electrons are shared equally.
The equation for the reaction is:
C₄H₈O₂ + C₂H₅OH = C₆H₁₂O₂ + H₂O
Now you see that the number of the moles of butanoic acid
and etyl butyrate is equal in
the reaction. That means;
number of moles of C₄H₈O₂ = number of moles of C₆H₁₂O₂
mass of C₄H₈O₂/ Molar mass of C₄H₈O₂ = mass of C₆H₁₂O₂/ molar mass of C₆H₁₂O₂
mass of C₆H₁₂O₂ = molar mass of C₆H₁₂O₂ x mass of C₄H₈O₂/ Molar mass of C₄H₈O₂
Now, assuming <span>100% yield, the mass
of ethyl butyrate produced is: </span>
<span>= 7.45/88.11 x 116.16</span>
<span>=9.82g</span>
<span>Thus, the theoretical yield of ethyl butyrate is 9.82g.</span>