In the reaction 2k + 2h2o → 2koh + h2, which compound has an element ratio of 2 : 1? h2o h2 k koh

1 answer:

4 0

The compound that has the element ratio 2:1 is 2H20.
In 2H2O, four atoms of hydrogen react with two hydrogen of oxygen to form molecules of water. Four divided by two will give two, that implies ratio 2;1. This implies that, two atoms of hydrogen will react with one atom of oxygen to give a molecule of water.

You might be interested in

Donald is writing a chemical name from a chemical formula. He has just identified that there are five atoms of oxygen bonded wit

gtnhenbr [62]

P2O5
hope this helps:)))))

4 0

3 years ago

Read 2 more answers

The ____________ in a chemical reaction is completely used up.

soldi70 [24.7K]

The answer is A reactant

7 0

3 years ago

Charina says that when waves interact with an object,

timama [110]

C I just took that quic

Explanation:

7 0

3 years ago

Read 2 more answers

N2+3H2---> 2NH3

irakobra [83]

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!

4 0

3 years ago

What is the mass in grams of 150 x 10^12 iron atoms, each having a mass of 56 u?

ad-work [718]

If we divide the number of Fe atoms (150x10^12) by Avogadro's Number (6.022x10^23) we get the number of moles of Iron. This value comes out to be 2.49x10^-10. If we multiply the number of moles by the molar mass of iron (55.86 grams) than we get a value that comes out to be 1.39x10^-8 grams or 13.9 nanograms.

5 0

4 years ago

Read 2 more answers