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3 years ago

N2 + 3H2 mc011-1.jpg 2NH3 What is the mole ratio of hydrogen to ammonia?

2 answers:

hjlf3 years ago

3 0

1 N₂ + 3 H₂ = 2 NH₃

Mole ratio of hydrogen to ammonia :

3 moles H₂ : 2 moles NH₃ or 3 : 2

hope this helps!

Vedmedyk [2.9K]3 years ago

3 0

Answer : The mole ratio of hydrogen to ammonia is, 3 : 2

Explanation :

Balanced chemical reaction : The chemical reaction in which the number of individual atoms of an element present on the reactant side always be equal to the number of individual atoms of an element present on the product side.

The given balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced chemical reaction, we conclude that 1 mole of $N_2$ gas react with 3 moles of $H_2$ gas to give 2 moles of $NH_3$ gas as a product.

The mole ratio of nitrogen, hydrogen and ammonia is, 1 : 3 : 2

Therefore, the mole ratio of hydrogen to ammonia is, 3 : 2

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Answer:

Size and Temperature or E & B

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2 years ago

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2.822 mol of nickel has _________ atoms of nickel. (Enter just the number for this one.)

Irina-Kira [14]

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To find the number of atoms in 2.822 moles of nickel, we need to multiply it by Avogadro's number. Avogadro's number is 6.02 x 10^23 atoms.

2.822 moles x (6.02 x 10^23) ≈ 1.698844 x 10^24

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3 years ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

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3 years ago

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konstantin123 [22]

Answer:false

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3 years ago

What is the bond between Ca and Cl

gtnhenbr [62]

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3 years ago