1) Convert the following into a balanced equation: when lead(II) nitrate is adde

Question

1) Convert the following into a balanced equation:

when lead(II) nitrate is added to sodium bromide solution, solid lead(II) bromide forms and sodium nitrate solution remains. Be sure to include the state of each reactant and product in the equation.

2) Calculate the maximum numbers of moles and grams of iodic acid (HIO3) that can form when 593g of iodine trichloride reacts with 136.7g of water.

ICl3+H2O--->ICl+HIO3+HCl (balance equation first to solve problem)

(A) mol HIO3 and g HIO3

(B) what mass of the excess reactant remain?

3) Metal hydrides react with water to form hydrogen gas and metal hydroxide. for example

SrH2(s)+2H2O(l)--->Sr(OH)2(s)+2H2(g)

you wish to calculate the mass of hydrogen gas that can be prepared from 4.32g of SrH2 and 3.60g of H2O.

(A) how many moles of H2 can be produced from the given mass of SrH2?

(B) how many moles of H2 can be produced from the given mass of H2O?

(C) which is the limiting reactant?

(D) how many grams of H2 can be produced?

PLEASE SHOW WORK TO GET FULL POINTS

Explanation / Answer

Pb(NO3)2 (s)+ 2NaBr (s) -----------> PbBr2(s) + 2NaNO3(s),

2) ICl3 moles = 593/466.528 = 1.27, H2O moles = 136.7/18 = 7.5944

2ICl3 + 3H2O --------> ICl + HIO3 + 5HCl,

A) mol of HIO3 = 1.27/2 = 0.635, mass of HIO3 = 0.635 x 175.9 = 111.7 gm

B) excess water left = (136.7)-( 0.635x3 x18) = 102.41 gm

3) A) moles of SrH2 = (4.32/89.6358) = 0.0482, moles of H2 = ( 0.0482x2) = 0.0964

B) moles of water = 3.6/18 = 0.2 , moles of H2 = 0.2,

C) SrH2 is limiting reagent

D) H2 produced = ( 0.0964x 2) = 0.1928 gm

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