For each of the following unbalance equation, calculate how many moles of the se

Question

For each of the following unbalance equation, calculate how many moles of the second reactant would be required to react completely with 0.557 grams of the first reactant
A. Al + Br2 = AlBr3 B. Hg + HClO4 = Al2(SO4)3 + H2 C. K + P = K3P D. CH4 + Cl2 = CCl4 + 2H2O For each of the following unbalance equation, calculate how many moles of the second reactant would be required to react completely with 0.557 grams of the first reactant
A. Al + Br2 = AlBr3 B. Hg + HClO4 = Al2(SO4)3 + H2 C. K + P = K3P D. CH4 + Cl2 = CCl4 + 2H2O
A. Al + Br2 = AlBr3 B. Hg + HClO4 = Al2(SO4)3 + H2 C. K + P = K3P D. CH4 + Cl2 = CCl4 + 2H2O

Explanation / Answer

A.

From unbalanced equation,

moles of Al = 0.557/26.981 = 0.0206 mols

moles of Br2 = 0.206 mols

mass of Br2 required = 0.0206 x 159.808 = 3.30 g

Balanced equation,

2Al + 3Br2 ---> 2AlBr3

moles of Al = 0.557/26.981 = 0.0206 mols

moles of Br2 = 3 x 0.0206/2 = 0.031 mols

mass of Br2 required = 0.031 x 159.808 = 4.95 g

B. Incorrect reaction

C. From unbalanced equation,

moles of K = 0.557/39.0983 = 0.014 mols

moles of P = 0.014 mols

mass of P required = 0.014 x 30.974 = 0.434 g

Balanced equation,

3K + P ---> K3P

moles of K = 0.557/39.0983 = 0.014 mols

moles of P = 4.75 mols

mass of P required = 4.75 x 30.974 = 0.147 g

D. From unbalanced equation,

moles of CH4 = 0.557/16.04 = 0.035 mols

moles of Cl2 = 2 x 0.035 = 0.035 mols

mass of Cl2 required = 0.035 x 70.906 = 2.46 g

Balanced equation [without O2],

CH4 + 2Cl2 ---> CCl4 + 2H2O

moles of CH4 = 0.557/16.04 = 0.035 mols

moles of Cl2 = 2 x 0.035 = 0.07 mols

mass of Cl2 required = 0.07 x 70.906 = 4.92 g

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