Chemistry 101 problems 1. If 10.0 g of molten iron (II) oxide reacts with 40.0 g

Question

Chemistry 101 problems

1. If 10.0 g of molten iron (II) oxide reacts with 40.0 g of magnesium what is the mass of iron produced?
FeO + Mg -> Fe + MgO

2. If 5.00 moles of hydrogen gas and 1.50 moles of oxygen gas react, what is the limiting reactant and how many moles of water are produced?
2 H2 + O2 -> 2 H2O

3. How many mL of hydrogen iodine are produced from 125 mL of H2 at standard pressure and temperature?
H2 + l2 -> 2 Hl

4. What mass of manganese (II) chloride must react with sulfuric acid to release 49.5 mL of hydrogen chloride gas at standard temperature and pressure?

MnCl2 + H2SO4 -> MnSO4 + 2 HCl

5. How many grams of mercuric chloride must react to afford 5.11 grams of mercury?
Co + HgCl2 -> CoCl3 + Hg

Explanation / Answer

1. If 10.0 g of molten iron (II) oxide reacts with 40.0 g of magnesium what is the mass of iron produced?
FeO + Mg -> Fe + MgO

72g + 24.3g-->56g + 40.3g

(divide all terms by 7.20) ironoxide being limiting reagent

10g + 3.375g-->7.78g + 5.97g

2. If 5.00 moles of hydrogen gas and 1.50 moles of oxygen gas react, what is the limiting reactant and how many moles of water are produced?
2 H2 + O2 -> 2 H2O

2moles(H2) + 1mole(O2) -> 2moles( H2O)

hence 5moles of H2 gas will require 2.5moles of O2 gas so O2 is limiting so multiply with 1.5

3moles(H2) + 1.5mole(O2) -> 3moles( H2O)

3. How many mL of hydrogen iodine are produced from 125 mL of H2 at standard pressure and temperature?
H2 + l2 -> 2 Hl

125ml=0.125L H2 or 0.125L/22.4L*moles=0.00558moles

since twicw number of moles of HI are produced so

moles of HI=0.0112moles*22.4L/mole=0.25L or 250mL

4. What mass of manganese (II) chloride must react with sulfuric acid to release 49.5 mL of hydrogen chloride gas at standard temperature and pressure?

MnCl2 + H2SO4 -> MnSO4 + 2 HCl

49.5mL H2 gas=0.0495L/22.4L*mole=0.00221moles

Half of the moles of MnCl2 should react with H2SO4=0.0011moles

mass of MnCl2=0.0011moles*126g/mole=0.1386g

5. How many grams of mercuric chloride must react to afford 5.11 grams of mercury?
Co + HgCl2 -> CoCl3 + Hg

5.11g Hg=5.11g/200.6g*moles=0.0255moles

moles of Hg=moles of HgCl2=0.0255moles

mass of HgCl2=0.0255moles*271.6g/mole=6.93g

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