Question

A. A bottle of 12.0 M hydrochloric acid has only 36.2 mL left in it.

What will the HCl concentration be if the solution is diluted to 260.0 mL ? M

B.

A flask containing 500 mL of 0.550 M HBr was accidentally knocked to the floor.

How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr(aq)+K2CO3(aq)→2KBr(aq)+CO2(g)+H2O(l)

Answer 1

A)

use dilution formula

M1*V1 = M2*V2

Here:

M1 is molarity of solution before dilution

M2 is molarity of solution after dilution

V1 is volume of solution before dilution

V2 is volume of solution after dilution

we have:

M1 = 12.0 M

V1 = 36.2 mL

V2 = 260.0 mL

we have below equation to be used:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (12*36.2)/260

M2 = 1.67 M

Answer: 1.67 M

B)

we have the Balanced chemical equation as:

2 HBr + K2CO3 ---> K2CO3 +

Molar mass of K2CO3 = 2*MM(K) + 1*MM(C) + 3*MM(O)

= 2*39.1 + 1*12.01 + 3*16.0

= 138.21 g/mol

From balanced chemical reaction, we see that

when 2 mol of HBr reacts, 1 mol of K2CO3 is reacted

mol of K2CO3 required = (1/2)* moles of HBr

= (1/2)*0.275

= 0.1375 mol

we have below equation to be used:

mass of K2CO3 = number of mol * molar mass

= 0.1375*1.382*10^2

= 19 g

Answer: 19.0 g