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sweet-ann [11.9K]
2 years ago
9

Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be

synthesized using a two-step thermal process. In the first step, phosphorus and oxygen react to form diphosphorus pentoxide: (l)(g)(g) In the second step, diphosphorus pentoxide and water react to form phosphoric acid: (g)(l)(l) Write the net chemical equation for the production of phosphoric acid from phosphorus, oxygen and water. Be sure your equation is balanced.
Chemistry
1 answer:
Sphinxa [80]2 years ago
6 0

Answer:

P_4_{(I)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}

Explanation:

The first step is:

P_4_{(I)}+5O_2_{(g)}\rightarrow 2P_2O_5_{(g)}

Second step is:

P_2O_5_{(g)}+3H_2O_{(l)}\rightarrow 2H_3PO_4_{(l)}

Multiplying second step by 2, and adding both the steps, we get that:

P_4_{(I)}+5O_2_{(g)}+2P_2O_5_{(g)}+6H_2O_{(l)}\rightarrow 2P_2O_5_{(g)}+4H_3PO_4_{(l)}

Cancelling common species, we get that:

P_4_{(l)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}

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The combustion of ethyne, shown below unbalance, produces heat which can be used to weld metals:
Andreyy89

Answer:

3.69 g

Explanation:

Given that:

The mass m = 325 g

The change in temperature ΔT = ( 1540 - 165)° C

= 1375 ° C

Heat capacity c_p = 0.490 J/g°C

The amount of heat required:

q = mcΔT

q =  325 × 0.490 × 1375

q = 218968.75 J

q = 218.97 kJ

The equation for the reaction is expressed as:

C_2H_{2(g)} + 5O_{2(g)} \to 2CO_{2(g)} + H_2O_{(g)}   \ \   \ \ \  \Delta H^o_{reaction} = -1544 \ kJ

Then,

1 mole of the ethyne is equal to 26 g of ethyne required for 1544 kJ heat.

Thus, for 218.97 kJ, the amount of ethyne gas required will be:

= \dfrac{26 \ g}{1544 \ kJ} \times 218.97 \ kJ

= 3.69 g

3 0
2 years ago
What is the maximum number of grams of ammonia, nh3, which can be obtained from the reaction of 10.0 g of h2 and 80.0 g of n2? n
Lelechka [254]

<u>Answer:</u> The mass of ammonia produced is 28.22 g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol

  • <u>For nitrogen gas:</u>

Given mass of nitrogen gas = 80.0 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen gas}=\frac{80.0g}{28g/mol}=2.86mol

The given chemical equation follows:

N_2+3H_2\rightarrow 2NH_3

By Stoichiometry of the reaction:

3 moles of hydrogen gas reacts with 1 mole of nitrogen gas

So, 5 moles of hydrogen gas will react with = \frac{1}{3}\times 5=1.66mol of nitrogen gas

As, given amount of nitrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of hydrogen gas produces 1 mole of ammonia

So, 5 moles of hydrogen gas will produce = \frac{1}{3}\times 5=1.66moles of ammonia

Now, calculating the mass of ammonia from equation 1, we get:

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 1.66 moles

Putting values in equation 1, we get:

1.66mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(1.66mol\times 17g/mol)=28.22g

Hence, the mass of ammonia produced is 28.22 g

4 0
2 years ago
What is the mass of a sample of water containing 3.55×1022 molecules of h2o?
ad-work [718]
6,02×10²³ -------------- 18g
3,55×10²² --------------  xx=\frac{3,55*10^{22}*18g}{6,02*10^{23}}=10,61*10^{22-23}g=10,61*10^{-1}g=1,061g


6 0
2 years ago
Read 2 more answers
Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such
8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

  • 1 A = 1 C/s
  • The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
  • 1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
  • The molar mass of silver is 107.87 g/mol

The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g

5 0
2 years ago
The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s
denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0
2 years ago
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