N2+3H2---> 2NH3

1 answer:

4 0

Hello!

A) How many grams of hydrogen are necessary to react completely with 50.0g of nitrogen in the above reaction?

The balanced chemical reaction is the following one:

N₂(g) + 3H₂(g) → 2NH₃(g)

To calculate the amount of hydrogen necessary we will use the following conversion factor to go from grams of nitrogen to grams of hydrogen:

$50 gN_2* \frac{1 mol N_2}{28,013 g N_2}* \frac{3 moles H_2}{1 mol N_2}* \frac{2,016 g H_2}{1 mol H_2}=10,7950 g H_2$

So, 10,7950 grams of H₂ are required to react with 50 g of nitrogen

B) How many grams of ammonia are produced in the reaction from the previous problem?

The balanced chemical reaction again is the following one:

N₂ + 3H₂ → 2NH₃

To calculate the amount of ammonia produced we will use the following conversion factor to go from grams of nitrogen to grams of ammonia:

$50 gN_2* \frac{1 mol N_2}{28,013 g N_2}* \frac{2 moles NH_3}{1 mol N_2}* \frac{17,031 g NH_3}{1 mol NH_3}=60,7968 g NH_3$

So, 60,7968 grams of NH₃ are produced from 50 g of nitrogen

C) How many grams of silver chloride are produced from 5.0g of silver nitrate reacting with an excess of barium chloride?

The balanced chemical equation for the reaction is the following one:

2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

To calculate the mass of Silver Chloride produced we will use the following conversion factor to go from grams of Silver Nitrate to grams of Silver Chloride:

$5 g AgNO_3* \frac{1 mol AgNO_3}{169,87 g AgNO_3}* \frac{2 moles AgCl}{2 moles AgNO_3}* \frac{143,32 g AgCl}{1 mol AgCl}=4,2185 g AgCl$

So, 4,219 g of AgCl are produced from the reaction of 5 g of AgNO₃

D) How much barium chloride is necessary to react with the silver nitrate in the previous problem?

The balanced chemical equation for the reaction is the following one:

2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

To calculate the mass of Barium Chloride necessary we will use the following conversion factor to go from grams of Silver Nitrate to grams of Barium Chloride:

$5 g AgNO_3* \frac{1 mol AgNO_3}{169,87 g AgNO_3}* \frac{1 mol BaCl_2}{2 moles AgNO_3}* \frac{208,23 g BaCl_2}{1 mol BaCl_2}=3,08 g BaCl_2$

So, 3,08 g of BaCl₂ are produced from the reaction of 5 g of AgNO₃

Have a nice day!

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