Metallic copper is formed when aluminum reacts with copper(II) oxide sulfide. Ho
ID: 529478 • Letter: M
Question
Metallic copper is formed when aluminum reacts with copper(II) oxide sulfide. How many grams of metallic copper can be Which of the following would be the limiting reagents in the reaction shown below? 2H_2(g) + O_2(g) rightarrow 2H_2O(g) a. 50 molecules H_2 b. 50 molecules of O_2 c. Neither a is limiting d. Both a and b are considered limiting reagents What is the maximum of grams of PH_3 that can be formed with 6.2 g of phosphorous reacts with 4.0 g of hydrogen to form PH_3 P_4(g) + 6H_2(g) rightarrow 4PH_3(g) a. 270 g b. 0.43 g c. 6.8 g d. 45 g Lead nitrate can be decomposed by heating. What is the percent of the decomposition reaction of 9.0 g of Pb(NO_3)_2 are boasted to give 5.5 g of PbO? 2Pb(NO_3)_2(s) rightarrow 2PbO(s) + 4NO_2(s) +O_2(g) Hydrogen gas can be produced by reacting aluminum with sulfuric acid. How many moles of sulfuric acid are needed to completely react with 15.0 mol of aluminum? In a particular reaction between copper metal and silver nitate, 12.7 g Cu produced 38.1 g Ag. What is the percent yield of silver in this reaction? Cu + 2AgNO_3 rightarrow Cu(NO_3)_2 + 2Ag a. 77.3% b. 88.2% c. 56.7% d. 176% Solid sodium reacts violently with water, producing heat hydrogen gas, sodium hydroxide. How many molecules of hydrogen gas are formed when 48.7 sodium are added to water? 2Na + 2H_2O rightarrow 2NaOH + H_2 What is the mole ratio of D to A in the generic chemical reaction? 2A + B rightarrow C + 3D How many grams of CO are needed to react with an excess Fe_2O_3 to produce 209.7 g Fe? Fe_2O_3(s) + 3CO(g) rightarrow 3CO_2(g) + 2Fe(s) If 8.00 mol of NH_3 reacted with 14.0 mol of O_2, how many moles of H_2O will be produce? 4NH_3(g) + 7O_2(g) rightarrow 4NO_2 + 6H_2O(g)Explanation / Answer
20 is illegible; the values are difficult to read.
21) The given reaction is
2 H2 (g) + O2 (g) ------> 2 H2O (g)
As per the balanced stoichiometric reaction,
2 molecules of H2 = 1 molecule of O2
Therefore, 50 molecules of H2 = ½*50 = 25 molecules of O2
50 molecules of O2 = 2*50 = 100 molecules of H2
Therefore, both H2 and O2 can be limiting reactants, but it depends on the number of other molecules present. Suppose we have 50 molecules of H2 with only 20 molecules of O2, then O2 will be the limiting reactant. However, if we have 50 molecules of H2 with 150 molecules of O2, then H2 will be the limiting reactant.
Ans: (d)
22) The given reaction is
P4 (g) + 6 H2 (g) -------> 4 PH3 (g)
As per the balanced stoichiometric reaction,
1 mole P4 = 6 moles H2 = 4 moles PH3
Molar mass of P4 = (4*30.97) g/mol = 123.88 g/mol.
6.2 g of phosphorus = (6.2 g)/(123.88 g/mol) = 0.0500 mole.
Molar mass of H2 = (2*1.008) g/mol = 2.016 g/mol
4.0 g of hydrogen = (4.0 g)/(2.016 g/mol) = 1.9841 mole.
Find out the limiting reactant.
0.0500 mole phosphorus = (0.0500 mole P4)*(6 mole H2/1 mole P4) = 0.3 mole H2
Offcourse, P4 is the limiting reactant and the amount of PH3 formed will be decided by P4.
As per the stoichiometric reaction,
1 mole P4 = 4 moles PH3.
Therefore,
0.0500 mole P4 = (0.0500 mole P4)*(4 mole PH3/1 mole P4) = 0.200 mole PH3.
Molar mass of PH3 = (1*30.97 + 3*1.008) g/mol = 33.994 g/mol.
Amount of PH3 formed = (33.994 g/mol)*(0.200 mole) = 6.7988 g 6.8 g (ans).
Ans: (c)