Question

6. Vinegar is 5.00% (m/v) acetic acid (CH3COOH). How many milliliters of vinegar would you need to react with an excess of baking soda (NaHCO3) in order to generate 10.0 L of carbon dioxide gas at 1.05 atm and a temperature of 25.0°C? 7. Shown below is the aqueous reaction of KNOX, HCl, and SnCl2. 2 KNO2 (aq) + 6 HCl(aq) + 2 SnCl2 (aq) → 2 SnCla (aq) + N20 (g) + 3 H20 (1) + 2 KCl (aq) a. Is this a redox reaction? If so, which atom was reduced and which atom was oxidized. Support your answer by determining the oxidation states of the relevant atoms. b. If 50.0 g of KNO, is consumed in this reaction, how many liters of N O would be produced at STP? C. How many grams of SnCl2 would need to react to produce 1.6 L of N O gas at a pressure of 1.25 atm and a temperature of 320 K? d. Consider a 250 ml solution of 86 mM KNO. If all of the KNO, in this solution is consumed by this reaction, how many moles of water will be produced?

Answer 1

CH, COOH + NaHCO - CHOCOONa + CO2 + H2O Given :-> Euccess of NaHCO3 is used. Hence aretic acid is the limiting reagent . As per eq. 1 mole of CH2COOH Produce I mole of C02 Now, we have to find out no. of rooles of CO2 pooduced. NO of moles of CO2 => py FORT n = p = 1.05 X 10.0 RT 0.08205 x 298 D= 0.429 mole :.for 0.429 mole of CO2, 0.429 mole of CH3COOH is sequired. i concentsanon (m): no. of moles volume con NOW, 54 CMIN) = 58 of CHCOOH + 95 mL of vio X / Vol (L) .. molanty = Amount of CHE COOH molao mass of ACOH (gimol) - 5 g 1 60 g moi o.ogs L concCM) = 0.877 M. i volume (L) = no. of moles of ACOH 0.429 conc. (m) 0. 877 = 0.489 L 489 mL of vinegas would need to deacts with an excess of baking soda (NaHCO3 ) to produce 10.0 LCOą.