Problems with stoichiometry?

Chemistry (stoichiometry problems)?

• I was wondering if someone could look at my answers for a few stoichiometry problems and tell me if i did them correctly or not. 1)How many moles of Mg are required to react w/ 2.0 moles of HCl acid (Mg + 2HCl -->MgCl_2 + H_2) my answer : 1 mole Mg 2)Find the mass of sugar required to produce 0.0812moles of CO_2 (C_6H_12O_6 -->2C_2H_6O +2CO_2 my answer: 5.16 grams 3) find the limiting reactant if 4.1g of Cr and 9.3g of Cl_2 are mixed and heated until they combine (2Cr +3Cl_2 --> 2CrCl_3 my answer: 9.3 of Cl_2 is the limiting reactant 4) determine the % yield for the reaction b/w 15.8g of NH_3 and excess oxygen when 21,8g of NO are recovered (4NH_3 +5O_2 ----> 4NO + 6H_2O my answer: 86.9% if you could even just tell my if one was correct that would be greatly appriacated thank you soo much ^^

• Answer:

  1. this is correct. The coefficients in the balanced equation can represent moles of reagents. In your balanced equation we see the ratio 1:2 for Mg to HCl - 2 moles of HCl reacts completely with 1 mole of Mg. 2. 1 mole C6H12O6 produces 2 mole CO2: C6H12O6 ---> 2 C2H6O + 2CO2 0.0821 mol CO2 x ( 1 mol C6H12O6 / 2 mol CO2) x (180.12 g C6H12O6 / 1 mol C6H12O6) = 7.39-g C6H12O6 3. There is a 2:3 ratio of Cr to Cl2. Lets first determine moles of each element: Cr = 4.1-g/52-g/mol = 0.0788 mol Cr-------------Cl2 = 9.3-g/71.0-g/mol = 0.131-mol Cl2 To use all of the Cr we will need: 0.0788 mol Cr x (3 mol Cl2 / 2 mol Cr) = 0.1182 mol Cl2. We have more than that so, Cr is limiting reagent. 4. Lets determine the quantity of NO that should be produced by reacting 15.8-g of NH3. 15.8-g NH3 x (1mol NH3 / 17.0-g NH3) x ( 4 mol NO / 4 mol NH3) x (30-g NO / 1 mol NO) = 27.9-g NO should be produced. Only 21.8-g was recovered so... 21.8-g / 27.9 x 100% = 78.1%