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

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.

$NaHCO_3_M_r = 22.99 + 1.008 + 12.011+ 3 \times 16.0= 84.01 g/mol$.

$2.1kg\ NaHCO_3 \times \frac{1000g}{kg} \times \frac{mol}{84.01g/mol} = 24.997\ mol$.

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

$PV= nRT \implies V= \frac{nRT}{P}= \frac{ 24.997\ mol \times 8.2507m^3\ atm \times 298.15K }{mol \times K \times 1.23 atm} = 49967\ m^3$

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

$49967\ m^3 \times\frac{L}{1000\ m^3} =49.97L\ CO_2\ produced$

Answer 2

Answer : The correct option is, (B) $4.98\times 10^2L$

Explanation :

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

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

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

$(1.23atm)\times V=25mole\times (0.0821L.atm/mole.K)\times (298K)$

$V=498L=4.98\times 10^2L$

Therefore, the volume of carbon dioxide gas produced is, $4.98\times 10^2L$