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?

Question

Asked on June 17, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-06-17T08:39:48.067Z

To determine the grams of chromite (FeCr_2O_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_2CrO_4 (from the replacement reaction) to find moles of K_2CrO_4. 4. Use the mole ratio from the balanced chemical equation to find moles of FeCr_2O_4. 5. Convert moles of FeCr_2O_4 to grams using its molar mass. First, calculate the molar masses: Molar mass of PbCrO_4: Pb: 207.2 g/mol Cr: 52.00 g/mol O: 16.00 g/mol Molar mass of PbCrO_4 = 207.2 + 52.00 + (4 × 16.00) = 323.2 g/mol Molar mass of FeCr_2O_4: Fe: 55.845 g/mol Cr: 52.00 g/mol O: 16.00 g/mol Molar mass of FeCr_2O_4 = 55.845 + (2 × 52.00) + (4 × 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 g. The paint is 0.511\% PbCrO_4 by mass. Mass of PbCrO_4 = 1000 g paint × 0.511 g PbCrO_4100 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 × 1 mol PbCrO_4323.2 g PbCrO_4 = 0.0158106 mol PbCrO_4 Step 3: Determine moles of K_2CrO_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_2CrO_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_2O_4 from the balanced equation. The balanced equation is: 4FeCr_2O_4(s) + 8K_2CO_3(aq) + 7O_2(g) 2Fe_2O_3(s) + 8K_2CrO_4(aq) + 8CO_2(g) From the equation, 4 moles of FeCr_2O_4 produce 8 moles of K_2CrO_4. The mole ratio is 4:8, or 1:2. Moles of FeCr_2O_4 = 0.0158106 mol K_2CrO_4 × 4 mol FeCr_2O_48 mol K_2CrO_4 = 0.0079053 mol FeCr_2O_4 Step 5: Convert moles of FeCr_2O_4 to grams. Mass of FeCr_2O_4 = 0.0079053 mol FeCr_2O_4 × 223.845 g FeCr_2O_41 mol FeCr_2O_4 = 1.7695 g FeCr_2O_4 Rounding to three significant figures, the mass of chromite needed is 1.77 g. The final answer is 1.77 g. Send me the next one 📸

Accepted Answer · 2026-06-17T08:39:48.067Z

To determine the grams of chromite (FeCr_2O_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_2CrO_4 (from the replacement reaction) to find moles of K_2CrO_4. 4. Use the mole ratio from the balanced chemical equation to find moles of FeCr_2O_4. 5. Convert moles of FeCr_2O_4 to grams using its molar mass. First, calculate the molar masses: Molar mass of PbCrO_4: Pb: 207.2 g/mol Cr: 52.00 g/mol O: 16.00 g/mol Molar mass of PbCrO_4 = 207.2 + 52.00 + (4 × 16.00) = 323.2 g/mol Molar mass of FeCr_2O_4: Fe: 55.845 g/mol Cr: 52.00 g/mol O: 16.00 g/mol Molar mass of FeCr_2O_4 = 55.845 + (2 × 52.00) + (4 × 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 g. The paint is 0.511\% PbCrO_4 by mass. Mass of PbCrO_4 = 1000 g paint × 0.511 g PbCrO_4100 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 × 1 mol PbCrO_4323.2 g PbCrO_4 = 0.0158106 mol PbCrO_4 Step 3: Determine moles of K_2CrO_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_2CrO_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_2O_4 from the balanced equation. The balanced equation is: 4FeCr_2O_4(s) + 8K_2CO_3(aq) + 7O_2(g) 2Fe_2O_3(s) + 8K_2CrO_4(aq) + 8CO_2(g) From the equation, 4 moles of FeCr_2O_4 produce 8 moles of K_2CrO_4. The mole ratio is 4:8, or 1:2. Moles of FeCr_2O_4 = 0.0158106 mol K_2CrO_4 × 4 mol FeCr_2O_48 mol K_2CrO_4 = 0.0079053 mol FeCr_2O_4 Step 5: Convert moles of FeCr_2O_4 to grams. Mass of FeCr_2O_4 = 0.0079053 mol FeCr_2O_4 × 223.845 g FeCr_2O_41 mol FeCr_2O_4 = 1.7695 g FeCr_2O_4 Rounding to three significant figures, the mass of chromite needed is 1.77 g. The final answer is 1.77 g. Send me the next one 📸

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?

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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?

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