Question

7.) how many grams of lead iodine are produced from 6.0 moles NaI according to the balanced equation given? Pb(NO3)2 + 2 NaI —> 2 NaNO3 + Pbl2

Answer 1

Pb(NO3)2 + 2 NaI ====> 2 NaNO3 + Pbl2

It's already balanced.

We focus only on NaI and Pbl2, so we need their molar mass. (Please use the periodic table to calculate the molar mass)

For NaI = 149.89 g/mol

For Pbl2 = 461.01 g/mol (We use this)

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We already have 6.0 moles of NaI. So,

2.0 moles of NaI ---------------------- 461.01 g Pbl2

6.0 moles of NaI -------------------- x

`x\text{ = }\frac{6.0\text{ moles x 461.01 g}}{2.0\text{ moles}}=\text{ 1383.03 g }`

We must write this in scientific notation:

`1383.03=1.38303x10^3=1.4x10^3g\text{ (approx.)}`

Answer: (A.) 1.4x10^3 g Pbl2