Respuesta :
The balance equation for the reaction between sodium bicarbonate and hydrochloric acid:
[tex]NaHCO_3_(_s_) + HCl_(_a_q_) \implies NaCl_(_a_q_) + CO_2_(_g_) + H_2O_(_l_)[/tex]
The reactants [tex]NaHCO_3[/tex] and [tex]HCl[/tex] react in the ratio 1:1. So we use the mass of sodium bicarbonate and molar mass to find the number of moles produced.
[tex]NaHCO_3_M_r = 22.99 + 1.008 + 12.011+ 3 \times 16.0= 84.01 g/mol[/tex].
[tex]2.1kg\ NaHCO_3 \times \frac{1000g}{kg} \times \frac{mol}{84.01g/mol} = 24.997\ mol[/tex].
Again we use the fact that the stochiometric ratio are 1:1:1:1:1, hence the moles of [tex]CO_2[/tex] are 24.977moles.
So we used the ideal gas law [tex]PV=nRT[/tex], where P is the pressure, V is the volume, R the gas constant, is [tex]8.2057m^3\ atm\ mol^-^1 K^-^1[/tex] and T is the temperature in kelvins. We make V the subject
[tex]PV= nRT \implies V= \frac{nRT}{P}= \frac{ 24.997\ mol \times 8.2507m^3\ atm \times 298.15K }{mol \times K \times 1.23 atm} = 49967\ m^3[/tex]
The answer is supposed to be in L, [tex]1L = 1000\ m^3[/tex], so
[tex]49967\ m^3 \times\frac{L}{1000\ m^3} =49.97L\ CO_2\ produced[/tex]
Answer : The correct option is, (B) [tex]4.98\times 10^2L[/tex]
Explanation :
First we have to calculate the moles of [tex]NaHCO_3[/tex].
[tex]\text{Moles of }NaHCO_3=\frac{\text{Mass of }NaHCO_3}{\text{Molar mass of }NaHCO_3}=\frac{2100g}{84g/mole}=25moles[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex].
The balanced chemical reaction will be,
[tex]NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O[/tex]
From the balanced chemical reaction, we conclude that
As, 1 mole of [tex]NaHCO_3[/tex] react to give 1 mole of [tex]CO_2[/tex]
So, 25 mole of [tex]NaHCO_3[/tex] react to give 25 mole of [tex]CO_2[/tex]
Now we have to calculate the volume of [tex]CO_2[/tex].
Using ideal gas equation,
[tex]PV=nRT[/tex]
where,
P = pressure of [tex]CO_2[/tex] gas = 1.23 atm
V = volume of [tex]CO_2[/tex] gas = ?
T = temperature of [tex]CO_2[/tex] gas = [tex]25^oC=273+25=298K[/tex]
n = number of moles of [tex]CO_2[/tex] 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
[tex](1.23atm)\times V=25mole\times (0.0821L.atm/mole.K)\times (298K)[/tex]
[tex]V=498L=4.98\times 10^2L[/tex]
Therefore, the volume of carbon dioxide gas produced is, [tex]4.98\times 10^2L[/tex]
