Pure dry nitrogen monoxide gas can be made reacting potassium nitrite, potassium

Question

Pure dry nitrogen monoxide gas can be made reacting potassium nitrite, potassium nitrate and chromium(IIII) oxide. The products are nitrogen monoxide and potassium chromate. How many grams of nitrogen monoxide can be produced from a reaction mixture containing 5.00 moles of each reactant? A mixture of competition 70.0% methane (CH_4) and 30% ethane (C_2H_6) by mass is burned in oxygen to produce carbon dioxide and water. The reactions that occur are CH_4 + 2O_2 rightarrow CO_2 + 2 H_2O 2 C_2H_6 + 7 O_2 rightarrow 4 CO_2 + 6 H_2O How many grams of oxygen are needed to react completely with 50.0 grams of the hydrocarbon mixture? 86 grams 105 grams 196 grams 294 grams A chemical reaction produces 1.77 moles of oxygen gas at 27 degree C and 0.933 atm. What volume in liters would this sample occupy? 40.6 L 35.5 L 46.7 L 38.7 L 39.6 L

Explanation / Answer

(33)

The raction can be written as,

Cr2O3 + 3 KNO2 + KNO3 ---------> 2 K2CrO4 + 4 NO

If each reactant is taken 5 moles, KNO2 would be the limiting reagent as it needs 3 moles for every 1 mol or remaining each reactant.

From the balancedequation,

3 mol KNO2 can form 4 mol NO

then, 5 mol KNO2 can form 5 * 4 / 3 = 6.67 mol of NO

Therefore, Mass of NO formed = 6.67 * 30 = 200 g.

(34)

Mass of methane = 50.0 * 70 / 100 = 35.0 g.

Mass of ethane = 50.0 * 30 / 100 = 15.0 g.

Moles of methane = 35.0 / 16 = 2.19 mol

Moles of ethane = 15.0 / 30.0 = 0.500 mol

Form the balanced equations provided,

2.19 mol of methane needs 2 * 2.19 = 4.38 mol of O2

0.500 mol ethane needs 0.500 * 7 / 2 = 1.75 mol of O2

Total moles of O2 needed = 4.38 + 1.75 = 6.13 mol

Mass of O2 neede = 6.13 * 32 = 196 g.

(c)

(35)

Ideal gas equation, P V = n R T
V = 1.77 * 0.0821 * 300 / 0.933

V = 46.7 L

(c)

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