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CuO 2 CHCOOH → Cu(CH2C00)2 + H20
1. "If a penny has 0.035 g of copper(II)oxide coating its surface, what volume of 0.10 M acetic acid
should be used to clean it?

Respuesta :

Answer:

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

sebassandin

Answer:

[tex]V=8.8x10^{-3}L=8.8mL[/tex]

Explanation:

Hello,

In this case, we balance the given chemical reaction as:

[tex]CuO_2+ 2CH_3COOH \rightarrow Cu(CH_3COO)_2 + 2H_2O[/tex]

In such a way, for the given 0.035 of copper (II) oxide, we compute the moles of acetic acid that will react with it considering their 1:2 mole ratio:

[tex]n_{CH_3COOH}=0.035gCuO*\frac{1molCuO}{79.545gCuO}*\frac{2molCH_3COOH}{1molCuO} =8.8x10^{-3}molCH_3COOH[/tex]

Hence, by using the formula of molarity we find the volume as:

[tex]M=\frac{n}{V}\rightarrow V=\frac{n}{M}=\frac{8.8x10^{-4}mol}{0.10mol/L} \\\\V=8.8x10^{-3}L=8.8mL[/tex]

Best regards.

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