To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses balanced chemical equations to calculate quantities of reactants or products.

Step 1: Calculate the molar mass of ethane (C2H6)
The molar mass of carbon (C) is approximately 12 g/mol and that of hydrogen (H) is approximately 1 g/mol. So, the molar mass of ethane is (2*12) + (6*1) = 30 g/mol.

Step 2: Calculate the number of moles of ethane
The number of moles of a substance is its mass divided by its molar mass. So, the number of moles of ethane is 180 g / 30 g/mol = 6 moles.

Step 3: Use the balanced chemical equation to find the number of moles of carbon dioxide
According to the balanced chemical equation, 1 mole of ethane produces 2 moles of carbon dioxide. So, 6 moles of ethane will produce 6 * 2 = 12 moles of carbon dioxide.

Step 4: Calculate the mass of carbon dioxide
The molar mass of carbon dioxide (CO2) is (1*12) + (2*16) = 44 g/mol. So, the mass of carbon dioxide is 12 moles * 44 g/mol = 528 g.

So, the mass of carbon dioxide formed when 180 g of ethane is burned in excess of oxygen is 528 g.

Question

To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses balanced chemical equations to calculate quantities of reactants or products.

Step 1: Calculate the molar mass of ethane (C2H6)
The molar mass of carbon (C) is approximately 12 g/mol and that of hydrogen (H) is approximately 1 g/mol. So, the molar mass of ethane is (2*12) + (6*1) = 30 g/mol.

Step 2: Calculate the number of moles of ethane
The number of moles of a substance is its mass divided by its molar mass. So, the number of moles of ethane is 180 g / 30 g/mol = 6 moles.

Step 3: Use the balanced chemical equation to find the number of moles of carbon dioxide
According to the balanced chemical equation, 1 mole of ethane produces 2 moles of carbon dioxide. So, 6 moles of ethane will produce 6 * 2 = 12 moles of carbon dioxide.

Step 4: Calculate the mass of carbon dioxide
The molar mass of carbon dioxide (CO2) is (1*12) + (2*16) = 44 g/mol. So, the mass of carbon dioxide is 12 moles * 44 g/mol = 528 g.

So, the mass of carbon dioxide formed when 180 g of ethane is burned in excess of oxygen is 528 g.

Knowee AI · Accepted Answer

To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses balanced chemical equations to calculate quantities of reactants or products.

Step 1: Calculate the molar mass of ethane (C2H6)
The molar mass of carbon (C) is approximately 12 g/mol and that of hydrogen (H) is approximately 1 g/mol. So, the molar mass of ethane is (2*12) + (6*1) = 30 g/mol.

Step 2: Calculate the number of moles of ethane
The number of moles of a substance is its mass divided by its molar mass. So, the number of moles of ethane is 180 g / 30 g/mol = 6 moles.

Step 3: Use the balanced chemical equation to find the number of moles of carbon dioxide
According to the balanced chemical equation, 1 mole of ethane produces 2 moles of carbon dioxide. So, 6 moles of ethane will produce 6 * 2 = 12 moles of carbon dioxide.

Step 4: Calculate the mass of carbon dioxide
The molar mass of carbon dioxide (CO2) is (1*12) + (2*16) = 44 g/mol. So, the mass of carbon dioxide is 12 moles * 44 g/mol = 528 g.

So, the mass of carbon dioxide formed when 180 g of ethane is burned in excess of oxygen is 528 g.

Calculate mass of carbon dioxide (CO2) formed when 180 g of ethane (C2H6) is burned in excess of oxygen.C2H6(g) + 3.5O2 (g) → 2CO2 (g) + 3H2O(l)Select the correct option from below.528 g880 g264 g528I DON'T KNOWSUBMIT ANSWER

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