1) Given the reaction of hydrogen and chlorine gases, calculate the moles of hyd

ID: 692405 • Letter: 1

Question

1) Given the reaction of hydrogen and chlorine gases, calculate the moles of hydrogen that react to produce 4.00 mol of HCl.
H2(g)+Cl2(g)2HCl(g)

2) Given the balanced equation, calculate the mass of product that can be prepared from 1.60 g of zinc metal.
2Zn(s)+O2(g)2ZnO(s)

3) How many grams of oxygen gas must react to give 2.30 g of ZnO?
2Zn(s)+O2(g)2ZnO(s)

4) Given the balanced equation, calculate the mass of product that can be prepared from 3.85 g of bismuth metal.
2Bi(s)+3Cl2(g)2BiCl3(s)

5) What is the mass of silver that can be prepared from 1.40 g of copper metal?
Cu(s)+2AgNO3(aq)Cu(NO3)2(aq)+2Ag(s)

6a) How many moles of chlorine gas react with 0.160 mol of metallic iron?

6b) How many moles of iron(III) chloride are produced?

H2(g)+Cl2(g)2HCl(g)

According to reaction 1 mole H2 produce 2 mole HCl therefore to produce 4 mole of HCl requied H2 = 2 mole

2 mole H2 requied

2Zn(s)+O2(g)2ZnO(s)

Zn molar mass = 65.38 gm/mole then 1.60 gm zn = 1.60/65.38 = 0.0245 mole

According to reaction 2 mole Zn produce 2 mole ZnO then 0.0245 mole of Zn produce 0.0245 ZnO

ZnO molar mass = 81.408 gm/mole then 0.0245 mole ZnO = 81.408 X 0.0245 = 1.99 gm of ZnO

1.99 gm product ZnO produced.

2Zn(s)+O2(g)2ZnO(s)

ZnO molar mass = 81.408 gm/mole then 2.30 gm ZnO = 2.30 / 81.408 = 0.02825 of ZnO

According to reaction 1 mole O2 produce 2 mole ZnO then to produce 0.02825 mole ZnO required O2 = 0.02825/2 = 0.01413 mole Of O2

molar mass of O2 = 32 gm /mole then 0.01413 mole = 0.01413 X 32 = 0.452 gm

0.452 gm O2 required

2Bi(s)+3Cl2(g)2BiCl3(s)

molar mass of Bi = 208.9804 gm per mole then 3.85 gm = 3.85 / 208.9804 = 0.0184 mole

According to reaction 2 mole Bi produce 2 mole BiCl3 then 0.0184 mole Bi produce 0.0184 mole BiCl3

molar mass of BiCl3 = 315.34 gm/mole then 0.0184 mole of BiCl3 = 0.0184 X 315.34 = 5.8 gm

5.8 gm BiCl3 produced.

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