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Sagot :
We need 8.8 mL of acetic acid
Explanation:
Step 1: Data given
Mass of Copper oxide (CuO) = 0.035 grams
Molar mass CuO = 79.545 g/mol
Molarity acetic acid = 0.10 M
Step 2: The balanced equation
CuO + 2CH3COOH → Cu(CH3COO)2 + H2O
Step 3: Calculate moles CuO
Moles CuO = mass CuO / molar mass CuO
Moles CuO = 0.035 grams / 79.545 g/mol
Moles CuO = 4.4 *10^-4 moles
Step 4: Calculate moles CH3COOH
For 1 mol CuO we need 2 moles CH3COOH to produce 1 mol Cu(CH3COO)2 and 1 mol H2O
For 4.4 *10^-4 moles CuO we'll have 2 * 4.4 * 10^-4 = 8.8 *10^-4 moles
Step 5: Calculate volume of acetic acid
Volume acetic acid = moles acetic acid / molarity
Volume acetic acid = 8.8 * 10^-4 moles / 0.10 M
Volume acetic acid = 0.0088 L = 8.8 mL
We need 8.8 mL of acetic acid
Explanation:
Step 1: Data given
Mass of Copper oxide (CuO) = 0.035 grams
Molar mass CuO = 79.545 g/mol
Molarity acetic acid = 0.10 M
Step 2: The balanced equation
CuO + 2CH3COOH → Cu(CH3COO)2 + H2O
Step 3: Calculate moles CuO
Moles CuO = mass CuO / molar mass CuO
Moles CuO = 0.035 grams / 79.545 g/mol
Moles CuO = 4.4 *10^-4 moles
Step 4: Calculate moles CH3COOH
For 1 mol CuO we need 2 moles CH3COOH to produce 1 mol Cu(CH3COO)2 and 1 mol H2O
For 4.4 *10^-4 moles CuO we'll have 2 * 4.4 * 10^-4 = 8.8 *10^-4 moles
Step 5: Calculate volume of acetic acid
Volume acetic acid = moles acetic acid / molarity
Volume acetic acid = 8.8 * 10^-4 moles / 0.10 M
Volume acetic acid = 0.0088 L = 8.8 mL
We need 8.8 mL of acetic acid
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