Which statement about states of matter is correct? (1 point)

A sample of limestone (calcium carbonate, CaCO3) is heated at 950 K until it is completely converted to calcium oxide (CaO) and

pav-90 [236]

Answer:

Therefore, volume of CO₂ produced in the first step is 9141.404 L

Explanation:

Equations of reactions:

A: CaCO₃(s) ---> CaO(s) + CO₂(g)

B: CaO(l) + H₂O(l) ---> Ca(OH)₂(s)

Molar mass of CaCO₃ = 100 g; molar mass of CaO = 56 g; molar mass of CO₂ = 44 g molar mass of H₂P = 18 g; molar mass of Ca(OH)₂ = 74 g

From equation B, 1 mole of CaO produces 1 mole of Ca(OH)₂

This means that 56 g of CaO produces 74 g of Ca(OH)₂

mass of CaO that produces 8.47 kg or 8470 g of Ca(OH)₂ = 8470 g * 56/74 = 6409.73 g of CaO

Therefore, 6409.73 g of CaO were produced in reaction A

From reaction A, 1 mole of CaCO₃ produces 1 mole CaO and 1 mole of CO₂

Number of moles of CaO in 6409.73 g = 6409.73 g/56 g/mol = 114.46 moles

Therefore, 114.46 moles of CO₂ were produces as well.

Molar volume of gas at STP = 22.4 litres

Volume of CO₂ produced at STP = 114.46 * 22.4 L =2563.904 L

However, the above reaction took place at 950 K and 0.976 atm, therefore volume of CO₂ produced under these conditions are obtained using the general gas equation

Using P₁V₁/T₁ = P₂V₂/T₂

P₁ = 1.0 atm, V₁ = 2563.904 L, T₁ = 273 K, P₂ = 0.976 atm, T₂ = 950 K, V₂ = ?

V₂ = P₁V₁T₂/P₂T₁

V₂ = (1.0 * 2563.904 * 950)/(0.976 * 273)

V₂ = 9141.404 L

Therefore, volume of CO₂ produced in the first step is 9141.404 L

3 0

2 years ago

Determine the partial pressure and number of moles of each gas in a 15.75-L vessel at 30.0 C containing a mixture of xenon and n

lawyer [7]

Answer:

The Partial pressure of Xe and Ne will be 4.95 atm and 1.55 atm. The number of moles of Xe and Ne will be 3.13 and 0.981

Explanation:

Let the total pressure of the vessel= 6.5 atm and mole fraction of Xenon= 0.761

As we know,

$\chi_{Ne} + \chi_{Xe} = 1\\\chi_{Ne}= 1- 0.761\\\chi_{Ne}= 0.239$

According to Dalton's Law of partial pressure-

$P_i=\chi_i\times P_{total}$

Where,

$P_i=$ The pressure of the gas component in the mixture

$\chi_i=$ Mole fraction of that gas component

$P_t=$ The total pressure of the mixture

$P_{Xe}=(0.761)\times(6.5)\\P_{Xe}= 4.95 atm\\\\\\P_{Ne}=(0.239)\imes (6.5)\\P_{Ne}= 1.55 atm$

<u>Calculation: </u>

To calculate the number of moles,

PV=nRT

$n=\frac{PV}{RT}$

$n_{Xe}= \frac{4.95\times 15.75}{0.0821\times303 }\\ n_{Xe}= \frac{77.96}{24.87} \\n_{Xe}= 3.13\,mole \\\\\\n_{Ne}= \frac{1.55\times 15.75}{0.0821\times303 }\\\\n_{Ne}=\frac{24.41}{24.87}\\ n_{Ne}=0.981 \,mole$