
Concept explainers
(a)
Interpretation:
For 10 mg of each reactant, the limiting reactant and mass of each product needs to be determined.
Concept Introduction:
The mass of the product can be found out by multiplying number of moles to the molar mass.
(a)

Answer to Problem 36A
The limiting reagent is CO, and the mass of CH3OH produced is 0.01142 grams.
Explanation of Solution
Now, calculate the number of moles of hydrogen and CO;
From the molar ratio CO:H2 = 1:2
0.000357 mol of CO requires 2×0.000357 mol of H2.
Hydrogen is in excess, hence the limiting reagent is CO.
Limiting Reactant: CO
(b)
Interpretation:
For 10 mg of each reactant, the limiting reactant and mass of each product needs to be determined.
Concept Introduction:
The mass of the product can be found out by multiplying the number of moles to the molar mass.
(b)

Answer to Problem 36A
The limiting reagent is I2,and the mass of AlI3 produced is 0.0107 grams
.
Explanation of Solution
Molar ratio of Al:I2
=2:3
0.00003940 moles of iodine molecule required
Al : I2
0.00002627 0.00003940
Al is in excess and hence iodine is the limiting reagent.
Molar ration of Al to AlI2 =1:1
Moles of AlI3 = 0.0000263 mol
Limiting Reactant: I2
(c)
Interpretation:
For 10 mg of each reactant, the limiting reactant and mass of each product needs to be determined.
Concept Introduction:
The mass of the product can be found out by multiplying the number of moles to the molar mass.
(c)

Answer to Problem 36A
Limiting Reagent is HBr, and mass of CaBr2 = 0.01235 grams, Mass of H2O = 1.112 x 10-3 grams
Explanation of Solution
Molar ratio of Calcium hydroxide: HBr=1:2
0.000123 mol of HBr required 1/2×0.0000123 moles of calcium hydroxide
HBr : Ca(OH)20.000123 0.0000618
Hence, calcium hydroxide is in excess, limiting reagent is HBr.
Limiting Reactant: HBr
(d)
Interpretation:
For 10 mg of each reactant, the limiting reactant and mass of each product needs to be determined.
Concept Introduction:
The mass of the product can be found out by multiplying the number of moles to the molar mass.
(d)

Answer to Problem 36A
Limiting Reagent is H2PO4,and mass of CrPO4 = 0.0151 grams, Mass of H2 = 2.0621 x 10-4 grams.
Explanation of Solution
Molar ratio of Cr:H2PO4=1:1
0.000102 mol of H2PO4required 0.000102 moles of Cr.
Hence, calcium hydroxide is in excess, limiting reagent is HBr.
Limiting Reactant: H3PO4
Chapter 9 Solutions
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