What is the mass % of nitrogen in urea, in arginine, and in or-nithine? (b) How many grams of nitrogen can be excreted as urea when 135.2 g of ornithine is produced?

To determine the grams of chromite (FeCr$_2$O$_4$) needed, we will follow these steps:

1. Calculate the mass of lead(II) chromate (PbCrO$_4$) required for 1 kilogram of paint.
2. Convert the mass of PbCrO$_4$ to moles using its molar mass.
3. Use the mole ratio between PbCrO$_4$ and K$_2$CrO$_4$ (from the replacement reaction) to find moles of K$_2$CrO$_4$.
4. Use the mole ratio from the balanced chemical equation to find moles of FeCr$_2$O$_4$.
5. Convert moles of FeCr$_2$O$_4$ to grams using its molar mass.

First, calculate the molar masses: Molar mass of PbCrO$_4$: Pb: $207.2 \text{ g/mol}$ Cr: $52.00 \text{ g/mol}$ O: $16.00 \text{ g/mol}$ $Molar mass of PbCrO_4 = 207.2 + 52.00 + (4 \times 16.00) = 323.2 g/mol$ Molar mass of FeCr$_2$O$_4$: Fe: $55.845 \text{ g/mol}$ Cr: $52.00 \text{ g/mol}$ O: $16.00 \text{ g/mol}$ $Molar mass of FeCr_2O_4 = 55.845 + (2 \times 52.00) + (4 \times 16.00) = 223.845 g/mol$

Step 1: Calculate the mass of PbCrO$_4$ in 1 kg of paint. A kilogram of paint is $1000 \text{ g}$. The paint is $0.511\%$ PbCrO$_4$ by mass. $Mass of PbCrO_4 = 1000 g paint \times \frac{0.511 g PbCrO_4}{100 g paint} = 5.11 g PbCrO_4$

Step 2: Convert the mass of PbCrO$_4$ to moles. $Moles of PbCrO_4 = 5.11 g PbCrO_4 \times \frac{1 mol PbCrO_4}{323.2 g PbCrO_4} = 0.0158106 mol PbCrO_4$

Step 3: Determine moles of K$_2$CrO$_4$. The problem states that "Lead(II) ion then replaces the K$^+$ ion" to form PbCrO$_4$. This implies a 1:1 mole ratio between K$_2$CrO$_4$ and PbCrO$_4$. $Moles of K_2CrO_4 = Moles of PbCrO_4 = 0.0158106 mol K_2CrO_4$

Step 4: Determine moles of FeCr$_2$O$_4$ from the balanced equation. The balanced equation is: $4FeCr_2O_4(s) + 8K_2CO_3(aq) + 7O_2(g) \longrightarrow 2Fe_2O_3(s) + 8K_2CrO_4(aq) + 8CO_2(g)$ From the equation, $4$ moles of FeCr$_2$O$_4$ produce $8$ moles of K$_2$CrO$_4$. The mole ratio is $4:8$, or $1:2$. $Moles of FeCr_2O_4 = 0.0158106 mol K_2CrO_4 \times \frac{4 mol FeCr_2O_4}{8 mol K_2CrO_4} = 0.0079053 mol FeCr_2O_4$

Step 5: Convert moles of FeCr$_2$O$_4$ to grams. $Mass of FeCr_2O_4 = 0.0079053 mol FeCr_2O_4 \times \frac{223.845 g FeCr_2O_4}{1 mol FeCr_2O_4} = 1.7695 g FeCr_2O_4$ Rounding to three significant figures, the mass of chromite needed is $1.77 \text{ g}$.

The final answer is $\boxed{1.77 g}$.

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