fill in the missing words\nsodium bicarbonate is one of the compounds used as an antacid:\n$nahco_3(aq)+hcl(a…

fill in the missing words\nsodium bicarbonate is one of the compounds used as an antacid:\n$nahco_3(aq)+hcl(aq)\rightarrow nacl(aq)+co_2(g)+h_2o(l)$.\ncalculate the pressure of carbon dioxide gas produced in a closed container when 0.325 g of sodium bicarbonate completely reacts with hydrochloric acid. assume the volume of 0.95 l and the temperature of 37 degrees celsius, following these steps.\n0.325 g of sodium bicarbonate is equivalent to \n\nmoles of sodium bicarbonate.\n\nmoles of carbon dioxide are produced. using the ideal gas law calculate the volume of the gas, but first convert the temperature to kelvin: 37 degrees celsius is \n\nkelvin. the pressure of produced carbon dioxide is \n\natm.

fill in the missing words\nsodium bicarbonate is one of the compounds used as an antacid:\n$nahco_3(aq)+hcl(aq)\rightarrow nacl(aq)+co_2(g)+h_2o(l)$.\ncalculate the pressure of carbon dioxide gas produced in a closed container when 0.325 g of sodium bicarbonate completely reacts with hydrochloric acid. assume the volume of 0.95 l and the temperature of 37 degrees celsius, following these steps.\n0.325 g of sodium bicarbonate is equivalent to \n\nmoles of sodium bicarbonate.\n\nmoles of carbon dioxide are produced. using the ideal gas law calculate the volume of the gas, but first convert the temperature to kelvin: 37 degrees celsius is \n\nkelvin. the pressure of produced carbon dioxide is \n\natm.

Answer

Explanation:

Step1: Calculate moles of sodium bicarbonate

The molar - mass of $NaHCO_3$ is $M = 23 + 1+12 + 3\times16=84\ g/mol$. The number of moles of $NaHCO_3$, $n_{NaHCO_3}=\frac{m}{M}=\frac{0.325\ g}{84\ g/mol}\approx0.00387\ mol$.

Step2: Determine moles of carbon dioxide

From the balanced chemical equation $NaHCO_3(aq)+HCl(aq)\rightarrow NaCl(aq)+CO_2(g)+H_2O(l)$, the mole ratio of $NaHCO_3$ to $CO_2$ is 1:1. So, $n_{CO_2}=n_{NaHCO_3}=0.00387\ mol$.

Step3: Convert temperature to Kelvin

The conversion formula from Celsius to Kelvin is $T(K)=T(^{\circ}C)+273.15$. So, $T = 37+273.15 = 310.15\ K$.

Step4: Use the ideal - gas law to calculate pressure

The ideal - gas law is $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal - gas constant ($R = 0.0821\ L\cdot atm/(mol\cdot K)$), and $T$ is temperature. We want to find $P$, so $P=\frac{nRT}{V}$. Substituting $n = 0.00387\ mol$, $R = 0.0821\ L\cdot atm/(mol\cdot K)$, $T = 310.15\ K$, and $V = 0.95\ L$ into the formula: [ \begin{align*} P&=\frac{0.00387\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times310.15\ K}{0.95\ L}\ &=\frac{0.00387\times0.0821\times310.15}{0.95}\ atm\ &\approx0.104\ atm \end{align*} ]

Answer:

0.00387; 0.00387; 310.15; 0.104