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sukhopar [10]
3 years ago
14

What mass of KNO3 would be needed to produce 18.4 liters of oxygen gas, measured at 1.50 x 10^3 kPa and 15 degrees Celsius?

Chemistry
1 answer:
Brut [27]3 years ago
7 0

Answer:-

2328.454 grams

Explanation:-

Volume V = 18.4 litres

Temperature T = 15 C + 273 = 288 K

Pressure P = 1.5 x 10^ 3 KPa

We know universal Gas constant R = 8.314 L KPa K-1 mol-1

Using the relation PV = nRT

Number of moles of oxygen gas n = PV / RT

Plugging in the values

n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)

n = 11.527 mol

Now the balanced chemical equation for this reaction is

2KNO3 --> 2KNO2 + O2

From the equation we can see that

1 mol of O2 is produced from 2 mol of KNO3.

∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.

= 23.054 mol of KNO3

Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol

Mass of KNO3 = 23.054 mol x 101 gram / mol

= 2328.454 grams

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3 years ago
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What is the energy change if 84.0 g of calcium oxide (CaO) reacts with excess water in the following reaction?
pochemuha
   The energy change  if 84.0 g   of CaO  react  with  excess  water is  98KJ of heat is released.

calculation
heat  =  number of moles  x  delta H

delta H = - 65.2  Kj/mol

first find the number of  moles  of  CaO reacted

moles = mass/molar mass
the molar mass  of CaO =  40 +  16=  56 g/mol
mass = 84 g
moles therefore = 84 g/56 g/mol =1.5 moles

Heat is therefore =  1.5 moles  x -65.2 = - 97.8 Kj = -98 Kj

  since  sign is  negative  the   energy  is released 

6 0
3 years ago
Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature
weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}     ------>NO_{(g)}

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

\delta H^0__{R,T_2} = \delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'

where:

\delta H^0__{R} = enthalpy of reaction

{\delta C_p(T')} = the difference in the heat capacities of the products and the reactants.

∴

\delta H^0__{R,435K} = \delta H^0__{R,298.15K} + \int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'

= 1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

\delta H^0__{R,435K} ≅ 91383 J

6 0
3 years ago
Which atom has the smallest atomic radius between cesium, potassium, rubidium, and francium?
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6.02 kJ>mol. When a small ice cube at -10°C is put into a cup of water at room temperature, which of the following plays a gr
9966 [12]

Answer:

Heat transfer during melting of ice plays greater role in cooling of liquid water.

Explanation:

Temperature of ice = -10 °c

Temperature of water = 0 °c

When ice cube is dipped in to the water.the heat transfer

Q = m c ΔT

⇒ Q = 1 × 2.01 × 10

⇒ Q = 20.1 KJ

Heat transfer during melting of ice Q_{melt} = latent heat of ice

Latent heat of ice = 334 KJ

⇒ Q_{melt} = 334 KJ

Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.

Thus heat transfer during melting of ice plays greater role in cooling of liquid water.

8 0
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