What is the limiting reagent when 3.2 mol of SO2 react with 2.2 mol of O2 according to this equation: What is the amount of unreacted excess reagent left over. 2SO2(g) + O2(g) -> 2SO3(g) b. Calculate the maximum amount of product that can be formed in moles and liters of gas-

c. If only 1.2 mol of SO3 gas was produced, then what was the actual yield in liters and what was the percent yield-

linda: 3 ; laura: 15

step-by-step explanation:

note that linda + laura's age is 18, and that linda is 12 years younger than her sister.

let "linda" = x , "laura" = y:

x + y = 18

x = y - 12

first, plug in y - 12 for x in the first equation:

(y - 12) + y = 18

simplify.

2y - 12 = 18

isolate the variable (y). note the equal sign, what you do to one side, you do to the other. do the opposite of pemdas. first add, then divide.

2y - 12 (+12) = 18 (+12)

2y = 18 + 12

2y = 30

(2y)/2 = (30)/2

y = 30/2

y = 15

next, plug in 15 for y in the second equation (the other equation would work too).

x = y - 12

x = (15) - 12

x = 3

linda's age is 3, laura's age is 15.

~

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"