Calcium carbide reacts with water according to the balanced chemical equation: \text{CaC}_… — JAMB UTME

Calcium carbide reacts with water according to the balanced chemical equation:

\text{CaC}_{2(s)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{Ca(OH)}_{2(aq)} + \text{C}_2\text{H}_{2(g)}

32.0\text{ g}

\text{CaC}_2

is reacted with

21.6\text{ g}

\text{H}_2\text{O}

, calculate the mass of the excess reactant remaining unreacted upon completion of the reaction.

[\text{Relative atomic masses: } \text{Ca} = 40, \text{C} = 12, \text{O} = 16, \text{H} = 1]

Cevap: 3.6 g

Cevap

3.6 g

Converting initial masses to mole values yields 0.50 mol of CaC₂ and 1.20 mol of H₂O. Based on the 1:2 mole ratio in the balanced equation, 0.50 mol of CaC₂ requires 1.00 mol of H₂O to react completely. This leaves 0.20 mol of H₂O unreacted. Converting 0.20 mol of H₂O to mass gives 0.20 mol × 18 g/mol = 3.6 g of excess reactant remaining.

Adım Adım Çözüm

Calculate molar masses of CaC₂ and H₂O

Molar mass of CaC₂ = 64 g/mol; Molar mass of H₂O = 18 g/mol

Molar mass is required to convert given mass values into mole quantities.

Convert initial masses to moles

Moles of CaC₂ = 0.50 mol; Moles of H₂O = 1.20 mol

Stoichiometric relationships depend strictly on mole ratios rather than direct mass ratios.

Determine limiting and excess reactants using mole ratios

CaC₂ is the limiting reactant; H₂O is in excess

According to the balanced equation coefficient ratio (1:2), 0.50 mol of CaC₂ requires 1.00 mol of H₂O. Because 1.20 mol of H₂O is present, H₂O is in excess.

Calculate remaining unreacted mass of excess reactant

3.6 g of excess H₂O remaining

Unreacted moles of H₂O = 1.20 - 1.00 = 0.20 mol. Mass = 0.20 mol × 18 g/mol = 3.6 g.

Anahtar Kavram

Determining limiting and excess reagents in chemical reactions and calculating unreacted leftover mass using mole ratios from balanced equations.

Tahmini Süre:2m 0s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru