Some dry chemicals can be used to put out forest fires. Sodium hydrogen carbonate is one such chemical that upon heating decomposes into carbon dioxide to suffocate the fire as shown in the balanced chemical reaction below. 2NaHCO_(3)(s)xrightarrow (Delta )Na_(2)CO_(3)(s)+H_(2)O(g)+CO_(2)(g). Determine the volume of carbon dioxide gas produced when 7.0 moles of Sodium hydrogen carbonate decomposes completely at STP. 78 Liters 160 Liters 310 Liters none of the above

To determine the volume of carbon dioxide gas produced when 7.0 moles of sodium hydrogen carbonate (NaHCO3) decomposes completely at standard temperature and pressure (STP), we need to use the balanced chemical equation and the ideal gas law.<br /><br />Given information:<br />- Balanced chemical equation: 2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g)<br />- 7.0 moles of sodium hydrogen carbonate (NaHCO3)<br /><br />Step 1: Determine the stoichiometry of the reaction.<br />From the balanced chemical equation, we can see that 2 moles of sodium hydrogen carbonate (NaHCO3) produce 1 mole of carbon dioxide (CO2).<br /><br />Step 2: Calculate the moles of carbon dioxide produced.<br />Moles of CO2 = (Moles of NaHCO3) × (1 mole of CO2 / 2 moles of NaHCO3)<br />Moles of CO2 = 7.0 moles × (1 mole of CO2 / 2 moles of NaHCO3)<br />Moles of CO2 = 3.5 moles<br /><br />Step 3: Calculate the volume of carbon dioxide gas produced at STP.<br />At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters.<br />Volume of CO2 = (Moles of CO2) × (22.4 liters / 1 mole)<br />Volume of CO2 = 3.5 moles × (22.4 liters / 1 mole)<br />Volume of CO2 = 78.4 liters<br /><br />Therefore, the volume of carbon dioxide gas produced when 7.0 moles of sodium hydrogen carbonate decomposes completely at STP is 78 liters.<br /><br />The correct answer is 78 Liters.