1. How many moles of Si are there in a sample of Si that contains 1.82×10 24 ato
ID: 705949 • Letter: 1
Question
1. How many moles of Si are there in a sample of Si that contains 1.82×1024 atoms?
____moles
What amount of the excess reagent remains after the reaction is complete? grams
3. For the following reaction, 0.361 moles of chlorine gas are mixed with 0.189 moles of water.
What is the formula for the limiting reagent?
What is the maximum amount of hydrochloric acid that can be produced?
___moles
4. For the following reaction, 4.04 grams of oxygen gas are mixed with excess ammonia. The reaction yields 2.07 grams of nitrogen monoxide.
What is the percent yield of nitrogen monoxide ?
What is the FORMULA for the limiting reagent?Explanation / Answer
1) No of mol of Si = (1.82*10^24)/(6.023*10^23) = 3.022 mol
2) CS2(s) + 4Cl2(g) ----> CCL4(l) + 2SCl2(S)
No of mol of CS2 = w/M = 24.7/76.14 = 0.324 mol
No of mol of Cl2 = 95.9/71 = 1.35 mol
limiting reactant = CS2
no of mol of CCl4 formed = 0.324*1/1 = 0.324 mol
amount of CCl4 formed = 0.324*153.82 = 49.84 g
3) 2cl2 + 2H2O ---> 4HCl + o2
chlorine gas = 0.361 mol
water = 0.189 mol
limiting reactant = H2O
no of mol of HCl formed = 0.189*4/2 = 0.378 mol
4) 4NH3 + 5O2 ---> 4NO + 6H2O
no of mol of O2 taken = 4.04/32 = 0.126 mol
theoretical yield of NO = 0.126*(4/5)*30 = 3.024 g
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