A 826.76 g sample of BrCl is in a 6342.7 mL container. What is the pressure of this gas in torr at 325.77 °C?

55.5atm

Explanation:

Using ideal gas law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

Molar mass of BrCl = 79.9 + 35.5

= 115.4g/mol

* mole = 826.76 g ÷ 115.4g/mol

mole = 7.16mol

* Volume = 6342.7 mL = 6342.7/1000 = 6.3427L

* Pressure = ?

* Temperature = 325.77 °C = 325.77 + 273 = 598.77K

PV = nRT

P = nRT/V

P = (7.16 × 0.0821 × 598.77) ÷ (6.3427)

P = 351.98 ÷ 6.3427

P = 55.5atm

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

it would be d. henry moseley

hope this !

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