Given the following reaction, answer the questions below. Simply type the numeri

Question

Given the following reaction, answer the questions below. Simply type the numerical answer into each box below. Be aware of sig figs. Units are not needed as they are already included after each answer! 2NaCI + Pb(N0_3)_2 rightarrow PbCI_2 + 2NaN0_3 If you have 8.0 moles of NaCI, how many moles of PbCI_2 can you form? If you have 6.0 moles of Pb(N0_3)_2, how many moles of PbCI_2 can you form? [moles Pb(N03)2] moles PbCI_2 What amount of PbCI_2 will form? moles PbCI_2 What reactant produces the smallest amount of PbCI_2 and is therefore the limiting reactants? is the limiting reactant (type the name of the chemical - spelling counts!)

Explanation / Answer

The reaction is:

2NaCl + Pb(NO3)2 -------> PbCl2 + 2NaNO3

In the reaction 2 moles of NaCl gives 1 mole os PbCl2.

So, 8.0 moles of NaCl will give PbCl2 = 8.0/2 = 4.0 moles of PbCl2.

In reaction, 1 mole of Pb(NO3)2 gives 1 mole of PbCl2.

So, 6.0 moles of Pb(NO3)2 will give PbCl2 = 1*6.0 = 6.0 moles of PbCl2.

Molar mass of PbCl2 = 278 g/mol

Mass of PbCl2 produced = moles*molar mass = 6.0*278 = 1668.0 g

Sodium chloride (NaCl) is limiting reactant as 2 moles of NaCl are needed in reaction.

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