All categories

3 years ago

How many grams of H2 will react with 350 g of N2?

Chemistry

1 answer:

miss Akunina [59]3 years ago

6 0

Answer:

10.80

Explanation:

As per the equation, let us calculate the mole ratio. N2+3H2→2NH3. As per the equation one mole of nitrogen reacts with 1 mol of hydrogen.

In terms of mass. 28.01 g of nitrogen needs 3 mol of hydrogen or 6.048 g of hydrogen.

We can set up the ratio;

28.01 g of

needs

6.048 g of

1 g of

needs

6.048

28.01

g of

50.0 g of

needs

6.048

50.0

28.01

g of

10.80 g of

You might be interested in

Plzz help guysss!! <br>How long would it take you to travel 8 miles at a speed of 1 mile per hour?

GrogVix [38]

Answer:

ksjdhfbjkdfbvbdvb

Explanation:

6 0

3 years ago

Which of these changes would be classified as producing a chemical change a. freezing water b. mixing salt and water to make a s

trasher [3.6K]

Answer:

Option B , Option C and Option D

4 0

3 years ago

What is required for an electron to move between energy levels?

Bingel [31]

It required a fixed finite amount these zones are known as energy levels

4 0

3 years ago

Read 2 more answers

Let’s assume that you put a balloon into the freezer. Initially,the balloon had 3.0liters of gas at a pressure of 400kPa and was

EastWind [94]

Here, we should use combined gas law which can be derived from combined gas law, “PV=nRT”. Rearranging, we can get PV/T=nR. Then we can set the two states in the problem together to get

P1V1/T1 = P2V2/T2

Then just plug in and solve algebraically.

Hope this helps

6 0

3 years ago

Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196 g sample of a

kvv77 [185]

Answer:

The mass percent of Al(OH)₃ is 15.3%

Explanation:

The reaction is:

Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O

The excess acid is neutralized with a solution of sodium hidroxide, in the reaction:

NaOH + HCl = NaCl + H₂O

The total moles of HCl is:

$n_{HCl,total} =M_{HCl} *V_{HCl} =0.111*0.025=2.78x10^{-3} moles$

From the second titration, the moles of excess of HCl is:

$n_{HCl,excess} =n_{NaOH} =M_{NaOH} *V_{NaOH} =0.132*0.01105=1.46x10^{-3} moles$

The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:

$n_{HCl,reacts} =n_{HCl,total}-n_{HCl,excess} =2.78x10^{-3} moles-1.46x10^{-3} moles=1.32x10^{-3} moles$

The ratio between HCl and Al(OH)₃ is 3:1. The MW for aluminum hydroxide is 78 g/mol, thus:

$m_{Al(OH)3} =1.32x10^{-3} molesHCl*\frac{1molAl(OH)3}{3molesHCl} *\frac{78gAl(OH)3}{1molAl(OH)3} =0.03g$

The percentage of Al(OH)₃ is:

$Percentage-Al(OH)3=\frac{m_{Al(OH)3} }{m_{antiacid} } *100=\frac{0.03}{0.196} =15.3$ %

3 0

3 years ago