Question

how many moles nano3 would be produced from the complete reaction of 253 g na2cro4? pb(no3)2 + na2cro4 → pbcro4 + 2nano3 ? mol nano3

Explanation:

Step1: Calculate the molar mass of \(Na_2CrO_4\)

The molar mass of \(Na\) is \(22.99\ g/mol\), \(Cr\) is \(52.00\ g/mol\), and \(O\) is \(16.00\ g/mol\).

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Step2: Calculate the number of moles of \(Na_2CrO_4\)

Using the formula \(n=\frac{m}{M}\), where \(m = 253\ g\) and \(M = 161.98\ g/mol\)

$$n(Na_2CrO_4)=\frac{253\ g}{161.98\ g/mol}\approx1.56\ mol$$

Step3: Use the stoichiometric ratio from the balanced equation

From the balanced equation \(Pb(NO_3)_2+Na_2CrO_4
ightarrow PbCrO_4 + 2NaNO_3\), the ratio of \(Na_2CrO_4\) to \(NaNO_3\) is \(1:2\)

$$n(NaNO_3)=2\times n(Na_2CrO_4)$$

Substitute \(n(Na_2CrO_4) = 1.56\ mol\)

$$n(NaNO_3)=2\times1.56\ mol = 3.12\ mol$$

Answer:

\(3.12\)