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calculate the percent composition of a 35.4 g piece of silver (Ag) that combines completely with 22.3 g of CO to form AgCL

User Octano
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1 Answer

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Step-by-step explanation:

To calculate the percent composition of a 35.4 g piece of silver (Ag) that combines completely with 22.3 g of CO to form AgCL, we need to know the molar mass of Ag and CO.

Ag = 107.87 g/mol

CO = 28.01 g/mol

The balanced equation for the reaction is:

Ag + CO → AgCL

So, we have 35.4 g of Ag and 22.3 g of CO reacting to form AgCL.

First, we need to calculate the moles of Ag and CO.

moles of Ag = 35.4 g / 107.87 g/mol = 0.327 moles

moles of CO = 22.3 g / 28.01 g/mol = 0.796 moles

Now, we can use the moles of each component to find the percent composition of Ag and CO in AgCL.

% of Ag = (moles of Ag / (moles of Ag + moles of CO)) x 100% = (0.327 / (0.327 + 0.796)) x 100% = 29.3%

% of CO = (moles of CO / (moles of Ag + moles of CO)) x 100% = (0.796 / (0.327 + 0.796)) x 100% = 70.7%

Therefore, the percent composition of Ag in AgCL is 29.3% and the percent composition of CO in AgCL is 70.7%.

User Jitin
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