Question

How much CO2 (L) is produced when 2.10 kg of sodium bicarbonate reacts with excess hydrochloric acid at 25.0 °C and 1.23 atm? A)4.17 × 10-1 B)4.98 × 102 C)4.17 × 101 D)3.50 E)4.98 × 10-1

Answer 1

Answer : The correct option is, (B)
`4.98* 10^2L`

Explanation :

First we have to calculate the moles of
`NaHCO_3`.

`\text{Moles of }NaHCO_3=\frac{\text{Mass of }NaHCO_3}{\text{Molar mass of }NaHCO_3}=(2100g)/(84g/mole)=25moles`

Now we have to calculate the moles of
`CO_2`.

The balanced chemical reaction will be,

`NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O`

From the balanced chemical reaction, we conclude that

As, 1 mole of
`NaHCO_3` react to give 1 mole of
`CO_2`

So, 25 mole of
`NaHCO_3` react to give 25 mole of
`CO_2`

Now we have to calculate the volume of
`CO_2`.

Using ideal gas equation,

`PV=nRT`

where,

P = pressure of
`CO_2` gas = 1.23 atm

V = volume of
`CO_2` gas = ?

T = temperature of
`CO_2` gas =
`25^oC=273+25=298K`

n = number of moles of
`CO_2` gas = 25 moles

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get

`(1.23atm)* V=25mole* (0.0821L.atm/mole.K)* (298K)`

`V=498L=4.98* 10^2L`

Therefore, the volume of carbon dioxide gas produced is,
`4.98* 10^2L`

Answer 2

The balance equation for the reaction between sodium bicarbonate and hydrochloric acid:

`NaHCO_3_(_s_) + HCl_(_a_q_) \implies NaCl_(_a_q_) + CO_2_(_g_) + H_2O_(_l_)`

The reactants
`NaHCO_3` and
`HCl` react in the ratio 1:1. So we use the mass of sodium bicarbonate and molar mass to find the number of moles produced.

`NaHCO_3_M_r = 22.99 + 1.008 + 12.011+ 3 * 16.0= 84.01 g/mol`.

`2.1kg\ NaHCO_3 * (1000g)/(kg) * (mol)/(84.01g/mol) = 24.997\ mol`.

Again we use the fact that the stochiometric ratio are 1:1:1:1:1, hence the moles of
`CO_2` are 24.977moles.

So we used the ideal gas law
`PV=nRT`, where P is the pressure, V is the volume, R the gas constant, is
`8.2057m^3\ atm\ mol^-^1 K^-^1` and T is the temperature in kelvins. We make V the subject

`PV= nRT \implies V= (nRT)/(P)= ( 24.997\ mol * 8.2507m^3\ atm * 298.15K )/(mol * K * 1.23 atm) = 49967\ m^3`

The answer is supposed to be in L,
`1L = 1000\ m^3`, so

`49967\ m^3 *(L)/(1000\ m^3) =49.97L\ CO_2\ produced`