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2 moles 4 moles 3 moles 1 mole

In the reaction: CH 4 + 2O 2  CO 2 + 2H 2 O How many moles of CH 4 react with 6 moles of O 2 ? . 2 moles 4 moles 3 moles 1 mole. How many moles of silver can be produced from 0.10 moles of tin and excess AgNO 3 in the following ( unbalanced) reaction?

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2 moles 4 moles 3 moles 1 mole

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  1. In the reaction:CH4 + 2O2 CO2 + 2H2O How many moles of CH4 react with 6 moles of O2? • 2 moles • 4 moles • 3 moles • 1 mole

  2. How many moles of silver can be produced from 0.10 moles of tin and excess AgNO3 in the following (unbalanced) reaction? Sn + AgNO3 Ag + Sn(NO3)2 . • .10 Moles • .20 Moles • .30 Moles • .40 Moles

  3. How many moles of chlorine gas are needed to react with 3.20 moles of sodium to produce sodium chloride? • 1.60 Moles Cl2 • 3.20 MolesCl2 • 6.40 Moles Cl2

  4. Hydrochloric acid reacts with calcium carbonate to produce H2CO3 and calcium chloride. How many moles of HCl are required to react completely with 25g of CaCO3? • .50 Mols HCl • 1.0 Mols HCl • .25 Mols HCl • .25 Mols HCl

  5. How many grams of aluminum chloride are produced when 32.0g of Al react with an excess of HCl? (Hydrogen gas is the other product) • 158 g of AlCl3 • 316 g of AlCl3 • 76.0 g of AlCl3

  6. What mass of zinc reacts completely with 250. mL of a .384M copper(II) nitrate solution? • 6.28 g • 12.56g • 3.14g

  7. When aluminum metal is placed in copper(II) nitrate, a single replacement reaction occurs. If the solution is made by dissolving 52.5g of copper(II) nitrate in water, and 2.7g of aluminum is placed in the solution: • How many grams of copper metal will be produced? • 19 g Cu • 9.5 g Cu • 18.0 g Cu • 18 g Cu

  8. How many grams of aluminum metal will be left over? • 1.0 g Al • 2.0 g Al • None • .50 g Al

  9. If 25.0g of zinc metal are reacted with 30.0g of lead(II) chloride, zinc chloride and lead metal are formed:What reactant will be in excess? • Zinc • PbCl2

  10. Calculate the mass of Zinc Chloride produced? • 14.2 g ZnCl2 • 14.7 g ZnCl2 • 14.0 g ZnCl2

  11. How many grams of the excess reactant will be left over? • 17.9 g • 16.5 g • 16.0 g • 12.2 g

  12. If 150. mL of a 0.10M solution of lead nitrate, Pb(NO3)2, is combined with 125 mL of a 0.20M solution of potassium iodide, KI What is the limiting reactant? • KI • B. Pb(NO33)2

  13. How many grams of the precipitate, lead iodide, will produced? • 2.5 g PbI2 • 5.8 g PbI2 • 2.8 g PbI2 • .70 g g PbI2

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