27. a hydrogen - filled balloon was ignited, and 1.50 g of hydrogen reacted with 11.9 g of oxygen. how many…

27. a hydrogen - filled balloon was ignited, and 1.50 g of hydrogen reacted with 11.9 g of oxygen. how many grams of water were formed? (assume that water vapour is the only product.)\n28. an automobile gasoline tank holds 21 kg of gasoline. when the gasoline burns, 84 kg of oxygen is consumed, and carbon dioxide and water are produced. what is the total combined mass of carbon dioxide and water that is produced?\n29. two samples of carbon tetrachloride were decomposed into their constituent elements. one sample produced 38.9 g of carbon and 448 g of chlorine, and the other sample produced 14.8 g of carbon and 134 g of chlorine. are these results consistent with the law of definite proportions? show why or why not.

27. a hydrogen - filled balloon was ignited, and 1.50 g of hydrogen reacted with 11.9 g of oxygen. how many grams of water were formed? (assume that water vapour is the only product.)\n28. an automobile gasoline tank holds 21 kg of gasoline. when the gasoline burns, 84 kg of oxygen is consumed, and carbon dioxide and water are produced. what is the total combined mass of carbon dioxide and water that is produced?\n29. two samples of carbon tetrachloride were decomposed into their constituent elements. one sample produced 38.9 g of carbon and 448 g of chlorine, and the other sample produced 14.8 g of carbon and 134 g of chlorine. are these results consistent with the law of definite proportions? show why or why not.

Answer

Explanation:

Step1: Write the chemical equation for hydrogen - oxygen reaction

The reaction is (2H_2 + O_2=2H_2O). First, find the moles of (H_2) and (O_2). The molar mass of (H_2) is (M_{H_2}=2\ g/mol), so the moles of (H_2), (n_{H_2}=\frac{1.50\ g}{2\ g/mol}=0.75\ mol). The molar mass of (O_2) is (M_{O_2} = 32\ g/mol), so the moles of (O_2), (n_{O_2}=\frac{11.9\ g}{32\ g/mol}\approx0.372\ mol).

Step2: Determine the limiting reactant

From the balanced equation, the mole - ratio of (H_2) to (O_2) is (2:1). For (0.75\ mol\ of\ H_2), the moles of (O_2) required is (\frac{0.75\ mol}{2}=0.375\ mol). Since (0.372\ mol\ of\ O_2) is present, (O_2) is the limiting reactant.

Step3: Calculate the moles of water formed

The mole - ratio of (O_2) to (H_2O) is (1:2). So the moles of (H_2O) formed, (n_{H_2O}=2\times n_{O_2}=2\times0.372\ mol = 0.744\ mol). The molar mass of (H_2O) is (M_{H_2O}=18\ g/mol), so the mass of (H_2O) formed, (m_{H_2O}=n_{H_2O}\times M_{H_2O}=0.744\ mol\times18\ g/mol = 13.4\ g).

Step4: Solve question 28 using the law of conservation of mass

According to the law of conservation of mass, the total mass of reactants equals the total mass of products. The mass of gasoline is (m_{gasoline}=21\ kg) and the mass of oxygen is (m_{O_2}=84\ kg). So the total mass of carbon dioxide and water produced is (m = 21\ kg+84\ kg=105\ kg).

Step5: Solve question 29 for the law of definite proportions

For the first sample, the ratio of chlorine to carbon is (\frac{m_{Cl_1}}{m_{C_1}}=\frac{448\ g}{38.9\ g}\approx11.52). For the second sample, the ratio of chlorine to carbon is (\frac{m_{Cl_2}}{m_{C_2}}=\frac{134\ g}{14.8\ g}\approx9.05). Since the ratios are not the same, the results are not consistent with the law of definite proportions.

Answer:

  1. (13.4\ g)
  2. (105\ kg)
  3. No, because the ratios of chlorine to carbon in the two samples are different.