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klasskru [66]
3 years ago
5

Nitric acid is produced commercially by the Ostwald process, represented by the following equations. 4 NH3(g) + 5 O2(g) 4 NO(g)

+ 6 H2O(g) 2 NO(g) + O2(g) 2 NO2(g) 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) What mass in kg of NH3 must be used to produce 7.6 ✕ 106 kg HNO3 by the Ostwald process, assuming 100% yield in each reaction?
Chemistry
1 answer:
Marrrta [24]3 years ago
4 0

Answer:

The answer to your question is: 1538095.2 kg of NH3

Explanation:

MW HNO3 = 63 kg

MW NO2 = 46 kg

                         3 NO2(g) + H2O(l)--- 2 HNO3(aq) + NO(g)

                            3(46) kg--------------   2(63) kg  

                                  x     ---------------  7600000 kg

           x = 7600000 x 138/126 = 8323809.5 kg og NO2

MW NO = 30            

                         2 NO(g) + O2(g)---2 NO2(g)

                       2(30) ------------------2(46)

                        x        ---------------- 8323809.5 kg  

             x = 8323809.5 x 60/92 = 5428571.4 kg of NO

MW NH3 = 17 kg

                      4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

                       4(17) -------------------- 4(30)

                         x ----------------------- 5428571.4

x = 5428571.4 x 34 / 120

x = 1538095.2 kg of NH3

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Which of the following is the BEST description of a budget? *<br> 10
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Given that you started with 28.5 g of K3PO4, how many grams of KNO3 can be<br> produced?
Irina18 [472]

Mass of KNO₃ : = 40.643 g

<h3>Further explanation</h3>

Given

28.5 g of K₃PO₄

Required

Mass of KNO₃

Solution

Reaction(Balanced equation) :

2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃

mol K₃PO₄(MW=212,27 g/mol) :

= mass : MW

= 28.5 : 212,27 g/mol

= 0.134

Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :

= 6/2 x mol K₃PO₄

= 6/2 x 0.134

= 0.402

Mass of KNO₃ :

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What is the Arrhenius definition of an acid
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A weather balloon is filled with helium that occupies a volume of 5.57 104 L at 0.995 atm and 32.0°C. After it is released, it r
Alchen [17]

6.52 × 10⁴ L. (3 sig. fig.)

<h3>Explanation</h3>

Helium is a noble gas. The interaction between two helium molecules is rather weak, which makes the gas rather "ideal."

Consider the ideal gas law:

P\cdot V = n\cdot R\cdot T,

where

  • P is the pressure of the gas,
  • V is the volume of the gas,
  • n is the number of gas particles in the gas,
  • R is the ideal gas constant, and
  • T is the absolute temperature of the gas in degrees Kelvins.

The question is asking for the final volume V of the gas. Rearrange the ideal gas equation for volume:

V = \dfrac{n \cdot R \cdot T}{P}.

Both the temperature of the gas, T, and the pressure on the gas changed in this process. To find the new volume of the gas, change one variable at a time.

Start with the absolute temperature of the gas:

  • T_0 = (32.0 + 273.15) \;\text{K} = 305.15\;\text{K},
  • T_1 = (-14.5 + 273.15) \;\text{K} = 258.65\;\text{K}.

The volume of the gas is proportional to its temperature if both n and P stay constant.

  • n won't change unless the balloon leaks, and
  • consider P to be constant, for calculations that include T.

V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}.

Now, keep the temperature at T_1 =258.65\;\text{K} and change the pressure on the gas:

  • P_1 = 0.995\;\text{atm},
  • P_2 = 0.720\;\text{atm}.

The volume of the gas is proportional to the reciprocal of its absolute temperature \dfrac{1}{T} if both n and T stays constant. In other words,

V_2 = V_1 \cdot\dfrac{\dfrac{1}{P_2}}{\dfrac{1}{P_1}} = V_1\cdot\dfrac{P_1}{P_2} = 4.72122\times 10^{4}\;\text{L}\times\dfrac{0.995\;\text{atm}}{0.720\;\text{atm}}=6.52\times 10^{4}\;\text{L}

(3 sig. fig. as in the question.).

See if you get the same result if you hold T constant, change P, and then move on to change T.

6 0
2 years ago
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