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PtichkaEL [24]
1 year ago
10

Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at

689 torr and 39 ∘C, is required to form 28 g of NaCl?
Chemistry
1 answer:
asambeis [7]1 year ago
6 0

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na +Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g(mass given of NaCl)/58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn,PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres

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jekas [21]

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<u><em>calculation</em></u>

 The new  volume is  calculated using  the Charles  law formula

that is  V1/T1= V2/T2

where T1= 25.0  c  into kelvin  =  25 +273  = 298 K

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Silver chloride is formed by mixing silver nitrate and barium chloride solutions. What volume of 1.50 M barium chloride solution
konstantin123 [22]

Answer:

1.22 mL

Explanation:

Let's consider the following balanced reaction.

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The molar mass of silver chloride is 143.32 g/mol. The moles corresponding to 0.525 g are:

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The molar ratio of AgCl to BaCl₂ is 2:1. The moles  of BaCl₂ are 1/2 × 3.66 × 10⁻³ mol = 1.83 × 10⁻³ mol.

The volume of 1.50 M barium chloride containing 1.83 × 10⁻³ moles is:

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