Calculating Amounts of Reactants and Products Determine how many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6: C6H12O6(s) + 6 O2(g)—»6 CO2(g) + 6 H2O(/) Sodium hydroxide reacts with carbon dioxide to form sodium carbonate and water: 2 NaOH(s) + CO2(g)—>Na2CO3(s) + H2O(/) How many grams of Na2CO3 can be prepared from 2.40 g of NaOH? (a) 3.18 g (b) 6.36 g (c) 1.20 g (d) 0.0300 g.
Answer : Step 1: In the above equation we came to know that one mole of glucose reacts with 6 moles of oxygen gas to give 6 molecules of carbon dioxide and 6 molecules of water . Step 2 : In the next step the sodium hydroxide reacts with carbon dioxide to give sodium carbonate and water. Here given that two moles of sodium hydroxide reacts with one mole of carbon dioxide to givr one mole of sodium carbonate and water. The mass of sodium hydroxide reacted is 2.40 g. Calculate the number of moles of sodium hydroxide as follows: Moles of NaOH= weight of NaOH/molar mass of NaOH. =2.40 g/40 g /mol. =0.06 mol.