Consider the following React ion. CH_4 (g) + O_2 (g) rightarrow CO_2 (g) + H_2O

Question

Consider the following React ion. CH_4 (g) + O_2 (g) rightarrow CO_2 (g) + H_2O (l) A chemist allows 23.2g of CH_4 and 78.3g O_2 to react. When the reaction is finished, the chemist collects 52.7g CO_2. Determine the limiting reagent, theoretical yield, and percent yield for the reaction. Consider the following React ion. Mg(OH)_2(aq) + H_3PO_4(l) rightarrow Mg_3(PO_)_2 (aq) + H_2O (l) A chemist allows 30.6 g of Mg(OH)_2 and 63.6g H_3PO_4 to react. When the reaction is finished, the chemist collects 34, 7g Mg_3(PO_4)_2. Determine the limiting reagent, theoretical yield, and percent yield for the reaction.

Explanation / Answer

balancing the given reaction is

CH4 + 2 O2 -----> CO2 + 2 H2O

mass of CH4 = 23.2 g

mass of O2 = 78.3 g

Mol. wt of CH4 = 16 g/mol

Mol. wt of O2 = 32 g/mol

No. of moles of CH4 = Mass / Mol. wt = 23.2 g / 16 g/mol = 1.45 moles

No. of moles of O2 = Mass / Mol. wt = 78.3 g / 32 g/mol = 2.45 moles

According to reaction stoichometry for 1 mole of CH4 we need 2 mol of O2

so for 1.45 moles of CH4 we need 2.9 moles of O2 but there is 2.45 moles of O2 is present.

so O2 is the limiting reactant. Answer

2 mole of O2 gives 1 mole of CO2

so 2.45 moles of O2 gives 1.225 moles of CO2

Mol. wt of CO2 = 44 g/mol

Mass of CO2 = 44 g/mol * 1.225 moles = 53.9 g

Theoretical yield = 53.9 g Answer

Given actual yield = 52.7 g

percent yield = ( actual yield / theoretical yield) * 100 % = 97.8 % Answer

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.