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defon
4 years ago
9

how many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia?

Chemistry
2 answers:
kkurt [141]4 years ago
7 0

Answer:

m_{(NH_4)_2SO_4}=7928.4g(NH_4)_2SO_4

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2NH_3+H_2SO_4-->(NH_4)_2SO_4

Now, by taking into account that 60.0 mol of sulfuric acid are used, the produced grams of ammonium sulfate, by applying the stoichiometric factors, turn out as shown below:

m_{(NH_4)_2SO_4}=60.0molH_2SO_4*\frac{1mol(NH_4)_2SO_4}{1molH_2SO_4} *\frac{132.14g(NH_4)_2SO_4}{1mol(NH_4)_2SO_4} \\m_{(NH_4)_2SO_4}=7928.4g(NH_4)_2SO_4

Best regards.

lianna [129]4 years ago
5 0
2NH3+H2SO4=>(NH4)2SO4
60(1/1)=60mol (NH4)2SO4
1mol=132.14grams
=7928.4grams

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Given 450.98 g of Cu(NO3)2, how many moles of Ag can be made? Provide your final answer rounded to two decimal places.
vivado [14]

Answer:

4.82 moles of Ag.

Explanation:

We'll begin by calculating the number of mole in 450.98 g of Cu(NO₃)₂. This can be obtained as follow:

Molar mass of Cu(NO₃)₂ = 63.5 + 2[14 + (16×3)]

= 63.5 + 2[14 + 48]

= 63.5 + 2[62]

= 63.5 + 124

= 187.5 g/mol

Mass of Cu(NO₃)₂ = 450.98 g

Mole of Cu(NO₃)₂ =?

Mole = mass /Molar mass

Mole of Cu(NO₃)₂ = 450.98 / 187.5

Mole of Cu(NO₃)₂ = 2.41 moles

Next, we shall determine the number of mole of Cu needed to produce 450.98 g (i.e 2.41 moles) of Cu(NO₃)₂. This can be obtained as follow:

Cu + 2AgNO₃ —> Cu(NO₃)₂ + 2Ag

From the balanced equation above,

1 mole of Cu reacted to produce 1 mole of Cu(NO₃)₂.

Therefore, 2.41 moles of Cu will also react to produce 2.41 moles of Cu(NO₃)₂.

Thus, 2.41 moles of Cu is needed for the reaction.

Finally, we shall determine the number of mole of Ag produced from the reaction. This can be obtained as follow:

From the balanced equation above,

1 mole of Cu reacted to produce 2 moles of Ag.

Therefore, 2.41 moles of Cu will react to produce = 2× 2.41 = 4.82 moles of Ag.

Thus, 4.82 moles of Ag were obtained from the reaction.

6 0
3 years ago
What mass of water is formed when 16g of hydrogen react with excess oxygen​
artcher [175]

Hello!

To start off, we must look at atomic masses. Atoms all have different weights, so we must first find hydrogen and oxygen's atomic masses.

Oxygen: 16.00 amu

Hydrogen: 1.01 amu

Now, moving on to the weight of water itself. Water has the formula of H20, with two hydrogen atoms and one oxygen. Therefore, <u>add up the amus to get the weight of one molecule of water.</u>

1.01 + 1.01 + 16.00 = 18.02 amu

Now, to see the ratio of each component. Since hydrogen weighs a total of 2.02 amu (1.01 + 1.01) in the entire atom, we can state that hydrogen makes up about 0.112 of the weight of water. Now apply that ratio to 16 g, and solve.

0.112x = 16

142.857143 = x

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