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3 years ago

Predict the overall reaction from the following two–step mechanism.

1 answer:

6 0

Mechanism
A2 = 2A fast
A + B --> AB slow

Rate = k[A2][B]

Rate = k[A2][B]1/2

Rate = k [A2]1/2[B]

Rate = k[A2]

Rate = k[A][B]
2A +B will become A2B,
but the reaction rate will be limited by the slowest step, here 2A becoming A2.
rate= k[A]^2
hope it helps

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3 years ago

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Answer:

There will be 525.2 grams of K3N produced

Explanation:

Step 1: Data given

Number of moles of potassium oxide ( K2O) = 6 moles

Magnesium nitride (Mg3N) = in excess

Molar mass of K3N = 131.3 g/mol

Step 2: The balanced equation

Mg3N2 + 3K2O → 3MgO + 2K3N

Step 3: Calculate moles of K3N

The limiting reactant is K2O.

For 1 mol Mg3N2 consumed, we need 3 moles of K2O to produce 3 moles of MgO and 2 moles of K3N

For 6 moles K2O we'll have 2/3 * 6 = 4 moles of K3N

Step 4: Calculate mass of K3N

Mass of K3N = moles K3N * molar mass K3N

Mass of K3N = 4 moles * 131.3 g/mol

Mass of K3N = 525.2 grams

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3 years ago

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3 years ago

Hydrobromic acid is added to a solution of lead(II) nitrate. Predict the products, and write a balanced equation for this reacti

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Answer:

2HBr + Pb(NO₃)₂ → PbBr₂ + 2HNO₃

Explanation:

Hydrobromic acid (HBr), reacts with lead (II) nitrate (Pb(NO₃)₂) producing Lead(II) bromide (PbBr₂) and nitric acid (HNO₃). The reaction is:

HBr + Pb(NO₃)₂ → PbBr₂ + HNO₃

This reaction is unbalanced. You can see in products 2 Bromides but in reactants just 1. And in reactants 2 NO₃ but as product just 1. Thus, you can balance the equation, thus:

2HBr + Pb(NO₃)₂ → PbBr₂ + 2HNO₃

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4 years ago

The rearrangement which most likely occurs in the following reaction can be described as a

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Answer:

b. hydride shift from C-3 to C-2.

Explanation:

Markovnikov's rule states that *in the addition of a protic acid HX or other polar reagent to an asymmetric alkene, the acid hydrogen (H) or electropositive part gets attached to the carbon with more hydrogen substituents, and the halide (X) group or electronegative part gets attached to the carbon with more alkyl substituents* (wikipedia).

This rule implies that the hydrogen of HBr will be attached to C-1 and the carbocation will be on C-2. Remember that the order of stability of carbocations is tertiary > secondary > primary > methyl. A hydride shift can yield a tertiary carbocation.

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3 years ago