ISM_chapter4_part2

ISM_chapter4_part2 - Chapter 4: Quantities of Reactants and...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 4: Quantities of Reactants and Products 137 58 . Answer: (a) 699 g (b) 526 g Strategy and Explanation: Given a balanced chemical equation and the mass of a reactant in kilograms, determine the mass of one of the products produced and the mass of another reactant required. Use metric conversion factors to convert kilograms to grams. Use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the product produced and the other reactant required. Then use molar masses to find the grams. 1.00 kg Fe 2 O 3 " 1000 g Fe 2 O 3 1 kg Fe 2 O 3 " 1 mol Fe 2 O 3 159.688 g Fe 2 O 3 = 6.26 mol Fe 2 O 3 (a) The balanced equation says: 1 mol Fe 2 O 3 produces 1 mol Fe. 6.26 mol Fe 2 O 3 " 2 mol Fe 1 mol Fe 2 O 3 " 55.845 g Fe 1 mol Fe = 699 g Fe (b) The balanced equation says: 1 mol Fe 2 O 3 requires 3 mol CO. 6.26 mol Fe 2 O 3 " 3 mol CO 1 mol Fe 2 O 3 " 28.0101 g CO 1 mol CO = 526 g CO Reasonable Answer Check: The mass of iron produced is less than the mass of iron(III) oxide it was produced from. The mass of CO used is less than the mass of iron(III) oxide even with a larger stoichiometric coefficient because CO contains lighter-weight atoms. 59. Answer: (a) K 2 PtCl 4 + 2 NH 3 Pt(NH 3 ) 2 Cl 2 + 2 KCl (b) 3.46 g K 2 PtCl 4 and 0.284 g NH 3 Strategy and Explanation: Given the reactants and products of a reaction and the desired mass of a product, balance the chemical equation and determine the mass of the reactants needed to produce that product. Given the formulas of the reactants and products, balance the equation, selecting appropriate order for the systematic balancing of all the atoms. Then use the molar mass of the product to find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the reactants needed to make that much product. Then use the molar mass of each to find the grams. (a) ? K 2 PtCl 4 + ? NH 3 ? Pt(NH 3 ) 2 Cl 2 + ? KCl Select order: N, H, Pt, K, Cl ? K 2 PtCl 4 + 2 NH 3 Pt(NH 3 ) 2 Cl 2 + ? KCl 2 N and 6 H K 2 PtCl 4 + 2 NH 3 Pt(NH 3 ) 2 Cl 2 + ? KCl 1 Pt K 2 PtCl 4 + 2 NH 3 Pt(NH 3 ) 2 Cl 2 + 2 KCl 2 K and 4 Cl (b) The balanced equation says: 1 mol Pt(NH 3 ) 2 Cl 2 requires 1 mol K 2 PtCl 4 . 2.50 g Pt(NH 3 ) 2 Cl 2 " 1 mol Pt(NH 3 ) 2 Cl 2 300.045 g Pt(NH 3 ) 2 Cl 2 = 0.00833 mol Pt(NH 3 ) 2 Cl 2 0.00833 mol Pt(NH 3 ) 2 Cl 2 " 1 mol K 2 PtCl 4 1 mol Pt(NH 3 ) 2 Cl 2 " 415.0866 g K 2 PtCl 4 1 mol K 2 PtCl 4 = 3.46 g K 2 PtCl 4 The balanced equation says: 1 mol Pt(NH 3 ) 2 Cl 2 requires 2 mol NH 3 . 0.00833 mol Pt(NH 3 ) 2 Cl 2 " 2 mol NH 3 1 mol Pt(NH 3 ) 2 Cl 2 " 17.0304 g NH 3 1 mol NH 3 = 0.284 g NH 3 Reasonable Answer Check: (a) Check the number of atom of each type in the reactants and products: 2 K, 1 Pt, 2 N, 6 H, and 4 Cl. The equation is properly balanced. (b) The mass of K 2 PtCl 4 is larger than the mass of Pt(NH 3 ) 2 Cl 2 , and the mass of NH...
View Full Document

This note was uploaded on 04/20/2008 for the course CHE 131 taught by Professor Kerber during the Spring '08 term at SUNY Stony Brook.

Page1 / 32

ISM_chapter4_part2 - Chapter 4: Quantities of Reactants and...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online