10. Given the balanced equation of NaOH(s)+CO_(2)(g)-CetNaHCO3 how many liters of Carboo Dioxide are needed to produce 2.00times 102g of NaHCO3 of at standard temperature and pressure? Sb_(2)O_(3) formed when you have 200 moles of oxygen gas? 2(88 moles 9.43 tnoles 53.3 tnoles 84.0 tnoles

To solve this problem, we need to determine how many liters of carbon dioxide (\(CO_2\)) are required to produce a given mass of sodium bicarbonate (\(NaHCO_3\)) at standard temperature and pressure (STP).<br /><br />### Step 1: Write the Balanced Chemical Equation<br /><br />The balanced chemical equation for the reaction is:<br /><br />\[ \text{NaOH} (s) + \text{CO}_2 (g) \rightarrow \text{NaHCO}_3 (s) \]<br /><br />### Step 2: Calculate Moles of \(NaHCO_3\)<br /><br />First, calculate the number of moles of \(NaHCO_3\) produced from the given mass.<br /><br />- Molar mass of \(NaHCO_3\):<br /> - Na: 22.99 g/mol<br /> - H: 1.01 g/mol<br /> - C: 12.01 g/mol<br /> - O: 16.00 g/mol × 3 = 48.00 g/mol<br /><br /> Total molar mass = 22.99 + 1.01 + 12.01 + 48.00 = 84.01 g/mol<br /><br />Given mass of \(NaHCO_3\) = \(2.00 \times 10^2\) g = 200 g<br /><br />Number of moles of \(NaHCO_3\):<br /><br />\[ \text{Moles of } NaHCO_3 = \frac{200 \text{ g}}{84.01 \text{ g/mol}} \approx 2.38 \text{ moles} \]<br /><br />### Step 3: Use Stoichiometry to Find Moles of \(CO_2\)<br /><br />From the balanced equation, the mole ratio of \(NaHCO_3\) to \(CO_2\) is 1:1. Therefore, 2.38 moles of \(NaHCO_3\) require 2.38 moles of \(CO_2\).<br /><br />### Step 4: Convert Moles of \(CO_2\) to Liters at STP<br /><br />At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.<br /><br />Volume of \(CO_2\) needed:<br /><br />\[ \text{Volume of } CO_2 = 2.38 \text{ moles} \times 22.4 \text{ L/mol} \approx 53.31 \text{ liters} \]<br /><br />Therefore, approximately 53.31 liters of carbon dioxide are needed to produce 200 grams of \(NaHCO_3\) at STP.