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

Answer:

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

Explanation:

$5NaBr+NaBrO_3 +3H_2SO_4\rightarrow

3Br_2+3Na_2SO_4+3H_2O

$

Moles of sodium bromide = $\frac{275 g}{103 g/mol}=2.6699 mol$

Moles of sodium bromate =$\frac{176 g}{151 g/mol}=1.1655 mol$

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:

$\frac{5}{1}\times 1.1655 mol=5.8278 mol$ 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:

$\frac{3}{5}\times 2.6699 mol=1.60194 mol$ 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

$PV=nRT$

$V=\frac{1.60194 mol\times 0.0821 atm L/ mol K\times 573 K}{0.9555atm}$

V = 78.87 L

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