I need help with 4 questions concerning chemistry. 1. For the following chemical

Question

I need help with 4 questions concerning chemistry.

1. For the following chemical reaction, how many moles of lead(II) iodide will be produced from 93.8 g of potassium iodide? 2KI + Pb(NO3)2 ---> PBI2 + 2KNO3

2. Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s). The equation for the reaction is 2KCIO3 ---> 2KCI +3O2  Calculate how many grams of O2(g) can be produced from heating 21.2 grams of KClO3(s).

3. In the following reaction, how many grams of sodium sulfate, Na2SO4, will be produced from 35.9 g of sodium hydroxide, NaOH? 2NaOH + H2SO4 ----> Na2SO4 + 2H2O

4. Given the following chemical equation, determine how many grams of N2 are produced by 9.17 g of H2O2 and 5.70 g of N2H4. 2H2O(I) + N2H4(I) --> 4H2O(g) + N2(g)

Thanks in advance!

Explanation / Answer

1).

1 mole of KI = 166 gms

=> 93.8 gm ok KI = = 0.56 moles of KI

2 moles KI produce 1 mole lead(II) iodide

=> 0.56 moles Ki will produce = 0.56/2 = 0.28 moles

2).

21.2 gms of KClO3 = 21.2/122.55 moles = 0.172 moles

and, 2 moles of KClO3 produce, 3 moles O2

=> 0.172 mol KClO3 produce = 3*0.172/2 = 0.258 moles O2

hence wt. of O2 produces = 0.258*32 = 8.25 gms

3).

35.9 gm NaOH = 35.9/40 moles = 0.897 moles

2 moles NaOH ==> 1 mole Na2SO4

=> 0.897 moles NaOH ==> 0.897/2 moles = 0.448 moles

and, 0.448 moles Na2SO4 = 142.04*0.448 = 63.63 gms

4).

9.17 gm of H2O2 = 9.17/20 moles = 0.459 moles

5.7 gm of N2H4 = 5.7/18 moles = 0.316 moles

2 moles of H2O2 reacts with 1 mole of N2H4

Hence H2O2 will exhaust first

2 moles of H2O2 ==> 1 mole of N2

=> 0.316 moles of H2O2 = 0.316/2 = 0.158 moles N2

=> 0.158 moles N2 = 0.158*14 gm N2 = 2.214 gm

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