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kolezko [41]
3 years ago
15

Phosphine, an extremely poisonous and highly reactive gas, will react with oxygen to form tetraphosphorus decoxide and water, as

described by the following UNBALANCED equation. PH3(g) + O2(g) → P4O10(s) + H2O(g) Calculate the mass of P4O10(s) formed when 225 g
Chemistry
1 answer:
erica [24]3 years ago
8 0

<u>Answer:</u> The amount of P_4O_{10} formed is 469.8 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     ......(1)

Given mass of phosphine = 225 g

Molar mass of phosphine = 34 g/mol

Putting values in equation 1, we get:

\text{Moles of phosphine}=\frac{225g}{34g/mol}=6.62mol

The given chemical reaction follows:

4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)

Assuming that oxygen gas is present in excess, it is considered as an excess reagent.

Phosphine is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of phosphine produces 1 mole of P_4O_{10}

So, 6.62 moles of phosphine will produce = \frac{1}{4}\times 6.62=1.655mol of P_4O_{10}

Now, calculating the mass of P_4O_{10} by using equation 1:

Molar mass of P_4O_{10} = 283.9 g/mol

Moles of P_4O_{10} = 1.655 moles

Putting values in equation 1, we get:

1.655mol=\frac{\text{Mass of }P_4O_{10}}{283.9g/mol}\\\\\text{Mass of }P_4O_{10}=(1.655mol\times 283.9g/mol)=469.8g

Hence, the amount of P_4O_{10} formed is 469.8 grams.

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Negatively charged particles in atom - Electrons
5 0
3 years ago
Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be
Sphinxa [80]

Answer:

P_4_{(I)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}

Explanation:

The first step is:

P_4_{(I)}+5O_2_{(g)}\rightarrow 2P_2O_5_{(g)}

Second step is:

P_2O_5_{(g)}+3H_2O_{(l)}\rightarrow 2H_3PO_4_{(l)}

Multiplying second step by 2, and adding both the steps, we get that:

P_4_{(I)}+5O_2_{(g)}+2P_2O_5_{(g)}+6H_2O_{(l)}\rightarrow 2P_2O_5_{(g)}+4H_3PO_4_{(l)}

Cancelling common species, we get that:

P_4_{(l)}+5O_2_{(g)}+6H_2O_{(l)}\rightarrow +4H_3PO_4_{(l)}

6 0
3 years ago
How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydr
andreyandreev [35.5K]

Answer:

percentage yield = 67%

Explanation:

Mass of Cu(NO₃)₂  = 15.25 g

Mass of NaOH   = 12.75 g

Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH   →  Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 15.25 g /187.56 g/mol

Number of moles = 0.08 mol

Moles of NaOH :

Number of moles = mass/ molar mass

Number of moles = 12.75 g / 40 g/mol

Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂.      NaOH             :      Cu(OH)₂

                               2                   :          1

                               0.32              :           1/2×0.32 = 0.16 mol

                            Cu(NO₃)₂         :           Cu(OH)₂

                                  1                  :               1

                             0.08                :              0.08

The number of moles produced by  Cu(NO₃)₂  are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield ×  100

percentage yield = 5.23 g/  7.808 g ×  100

percentage yield = 0.67 ×  100

percentage yield = 67%

5 0
3 years ago
2. How much heat is released when 432g of water cools down from 710C to 180C?
nadya68 [22]
The heat released by the water when it cools down by a temperature difference AT

is Q = mC,AT

where

m=432 g is the mass of the water

C, = 4.18J/gºC

is the specific heat capacity of water

AT = 71°C -18°C = 530

is the decrease of temperature of the water

Plugging the numbers into the equation, we find

Q = (4329)(4.18J/9°C)(53°C) = 9.57. 104J

and this is the amount of heat released by the water.
7 0
3 years ago
Which<br> countries were the world leaders in space exploration throughout the 1960's?
g100num [7]

Answer:

the us and the soviet union

Explanation:

7 0
3 years ago
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