Question

Cumulative problem. Consider the following balanced chemical reaction. How many liters of bromine gas (Bra) at 300 C and 735 torr are formed when 275 g of sodium bromide reacts with 176 g of sodium bromate (NaBrO)? (Hint! Find your limiting reactant... 5 NaBr(aq)+ NaBrO,(aq)+3 H,So(aq)3 Bralg)+3 Na,Sos(aq)+3 HOU)

Answer 1

78.87 liters of bromine gas at 300 °C and 735 Torr are formed.

Step-by-step explanation:

`5NaBr+NaBrO_3 +3H_2SO_4\rightarrow</p><p>3Br_2+3Na_2SO_4+3H_2O</p><p>`

Moles of sodium bromide =
`(275 g)/(103 g/mol)=2.6699 mol`

Moles of sodium bromate =
`(176 g)/(151 g/mol)=1.1655 mol`

According to reaction , 1 mol of sodium bromate reacts with 5 moles of sodium bromide. Then 1.1655 mol of sodium bromate will react with:

`(5)/(1)* 1.1655 mol=5.8278 mol` of sodium bromide.

This means that sodium bromide is in limiting amount the amount of bromine gas depends upon sodium bromide.

According to reaction 5 moles of sodium bromide gives 3 moles of bromine gas.

Then 2.6699 moles of sodium bromide will give:

`(3)/(5)* 2.6699 mol=1.60194 mol` of bromine gas

Volume occupied by bromine gas at 300 °C and 735 Torr.

Pressure of the gas = P =735 Torr = 0.9555 atm

Temperature of the gas = T = 300°C = 573 K

n = 1.60194 mol

`PV=nRT`

`V=(1.60194 mol* 0.0821 atm L/ mol K* 573 K)/(0.9555atm)`

V = 78.87 L

78.87 liters of bromine gas at 300 °C and 735 Torr are formed.