If 0.10 mole of sio2 and 0.20 mole hf are combined in the reaction sio2 + 4hf => sif4 + 2h2o which of these statements is false? a. hf is the limiting reactant. b. some sio2 will be left over. c. the theoretical yield of sif4 is 0.10 mole. d. more hf would be needed to ensure that all the reactants are used up. e. the theoretical yield of h2o will be 0.10 mole.

C. The Theoretical yield of SiF4 is = 0.10

Explanation:

A. is TRUE

0.10 mol SiO2  * 4 mol HF/mol SiO2 = 0.40 mol HF which we do not have. Hence, HF is the limiting reagent

B. is TRUE

Some Sio2 will be left over is true since our limiting reagent is HF

C.  is FALSE

0.020 mol HF * 1 mol SiF4 produced/ 4 mol HF = 0.05 mol SiF4 produced

Since the mol of SiF4 produced by  the limiting reagent HF is given in the ratio of 1/4  it follows that 0.05 mol SiF4 will be produced and no 0.10 therefore statement is FALSE

D.  is TRUE

Since HF is the limiting reagent is True that more HF will be needed to react completely with HF (exactly 0.20 mol HF) .

E.  is TRUE

The theoreical yield of wate is  0.10 mol TRUE

0.05 mol SiO2 * 2 mol H2O/ mol SiO2 = 0.10 mol H2O

the maximum mass of nh3 is nh3 in grams

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol