Question

name _ score / 100 warm up stoichiometry show your solution/work (no work / solution = zero). observe correct number of significant figures. encircle or box your final answer. answer on a separate sheet of paper. limiting reagent problem #1: for the combustion of sucrose: _na + _cl₂→ _nacl there are 3.00 g of cl₂ and 4.00 g of na reacting. which is the limiting reagent and excess reagent? molar mass of nacl: na = cl = g/mol nacl g cl₂ to mol cl₂ using atomic mass of cl₂ mol cl₂ to mol nacl using balanced eq mol nacl to g nacl using molar mass of nacl 7 g nacl = 3.00 g cl₂ x _ x _ x _ g na to mol na using atomic mass of na mol na to mol nacl using balanced eq mol nacl to g nacl using molar mass of nacl 7 g nacl = 4.00 g na x _ x _ x _ limiting reactant: _ excess reactant: _

Explanation:

Step1: Balance the equation

$2Na + Cl_2
ightarrow 2NaCl$

Step2: Calculate moles of $Na$

$n_{Na}=\frac{4.00\ g}{22.99\ g/mol}=0.174\ mol$

Step3: Calculate moles of $Cl_2$

$n_{Cl_2}=\frac{3.00\ g}{70.90\ g/mol}=0.0423\ mol$

Step4: Determine limiting reagent

From the balanced - equation, mole ratio of $Na$ to $Cl_2$ is 2:1. Moles of $Na$ needed for 0.0423 mol of $Cl_2$ is $2\times0.0423 = 0.0846\ mol$. Since we have 0.174 mol of $Na$, $Cl_2$ is the limiting reagent.

Step5: Determine excess reagent

$Na$ is the excess reagent.

Answer:

Limiting reagent: $Cl_2$
Excess reagent: $Na$