Ch. 11 Chem Test A

The amount of product formed when a reaction is carried out in the the laboratory is called the ______.

• actual yield

• theoretical yield

• limiting reactant

• excess reactant

The reactant that determines the amount of product that can be formed is thee ______________.

• actual yield

• theoretical yield

• limiting reactant

• excess reactant

_________ is a quantity of a reactant that has more than enough needed in order to react.

• actual yield

• theoretical yield

• limiting reactant

• excess reactant

The calculated amount of a product that might be formed during a reaction is the __________.

• actual yield

• theoretical yield

• stoichiometry

• limiting reactant

The calculation of quantities in chemical equations is called ________.

• actual yield

• theoretical yield

• stoichiometry

• excess reactant

The ratio of the actual yield to the theoretical yield is known as

• percent composition.

• percent yield.

• percent error.

• percent ratio.

Which of these expressions is an incorrect interpretation of the balanced equation below?

 $2S\ +\ 3SO_2\ \rightarrow\ 2SO_3$  

•  $2\ atoms\ S\ +\ 3\ molecules\ O_{2\ }\rightarrow\ 2\ molecules\ SO_3$  

•  $2\ g\ S\ +\ 3\ g\ O_2\rightarrow\ 2\ g\ SO_3$  

•  $2\ mol\ S\ +\ 3\ mol\ O_2\rightarrow\ 2\ mol\ SO_3$  

• all of these are incorrect

In a chemical reaction, the mass of the products

• is less than the mass of the reactants.

• is greater than the mass of the reactants.

• is equal to the mass of the reactants.

• has no relationship to the mass of the reactants.

A reaction that has been calculated to produce 60.0 g of  $CuCl_2$  actually produces 50.0 g of  $CuCl_2.$  What is the percent yield?

• 0.833%

• 83.3%

• 96.1%
  

• 120%

The equation for the complete combustion of methane is:

 $CH_4+2O_2\rightarrow CO_2+2H_2O$  
To calculate the number of grams of  $CO_2$  produced by the reaction of 29.5 g of  $CH_4$  with  $O_2$  , the first conversion factor to us is

•  $\frac{1\ mol\ CH_4}{16.0\ g\ CH_4}$  

•  $\frac{16.0\ g\ CH_4}{1\ mol\ CH_4}$  

•  $\frac{2\ mol\ O_2}{1\ mol\ O_2}$  

•  $\frac{29.5\ g\ CH_4}{2\ mol\ CO_2}$