What is the theoretical field of PH_3 for the reaction of 182 g of calcium phosp
ID: 538033 • Letter: W
Question
What is the theoretical field of PH_3 for the reaction of 182 g of calcium phosphide with 126 g of water? A 4.55 g sample of aluminum reacted completely with a 9.62 g sample of Nickel(II) oxide. After the reaction, which reactant was still present? What A mass of this reactant was present? A 15.8 g sample of magnesium was mixed with 14.6 g of hydrochloric acid and allowed to react completely. A 15.0 g sample of magnesium chloride was recovered. Calculate the percent yield. What mass of sulfur dichloride was needed to react with excess sodium fluoride to prepare 1.19 g of disulfur dichloride if the percent yield was 51.96? The reaction by-products are sulfur tetrafluoride and sodium chloride. A 0.10 L sample of C_6H_5NO_2 (d = 1.20 g/mL) was reacted with a 0.30 L sample of C6H1404 (d = 1.12 g/mL) and yielded 55 g of (C_6H_5N)_2. Determine the percent yield of this reaction. The by-products were C_6H_12O_4 and water.Explanation / Answer
4.
chemical reaction equation,
2Al + 3NiO ---> Al2O3 + 3Ni
moles Al = 4.55/27 = 0.17 mol
moles NiO = 9.62/74.7 = 0.13 mol
If all of Al is consumed we would need = 0.17 x 3/2 = 0.255 mol of NiO
Since NiO available is lower than needed, NiO is the limiting reactant
Reactant left at the end of reaction = Al
Amount of Al left at the end of reaction = 4.55 - 0.13 x 2 x 27/3 = 2.21 g
7.
Chemical reaction equation,
2C6H5NO2 + 4C6H14O4 ---> (C6H5N)2 + 4C6H12O4 + 4H2O
moles C6H5NO2 = 0.1 L x 1.2 x 1000/123.11 = 0.975 mol
moles C6H14O4 = 0.3 L x 1.12 x 1000/150.2 = 2.24 mol
If all of C6H5NO2 is consumed, we would need = 0.975 x 4/2 = 1.95 mol of C6H14O4
Since moles of C6H14O4 available is higher than needed, C6H5NO2 is the limiting reactant
Theoretical yield of (C6H5N)2 = 0.975 x 182.2/2 = 88.82 g
Percent yield = 55 x 100/88.82 = 61.92%
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