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2 years ago

How many grams of hydrogen are needed to react with 1.40 g of nitrogen to produce ammonia?

Chemistry

1 answer:

vazorg [7]2 years ago

8 0

Answer:

Write a balanced chemical reaction:

N2 + 3H2 ==> 2NH3

Looking at the mole ratios in this balanced equation you can see it takes 3 moles H2 to make 2 moles NH3. So, next calculate the moles of NH3 represented by 1.80 g and then convert to moles of H2 needed:

moles of NH3 = 1.80 g x 1 mole/17 g = 0.106 moles NH3

Moles H2 needed = 0.106 moles NH3 x 3 moles H2/2 moles NH3 = 0.159 moles H2 needed

Grams H2 needed = 0.159 moles x 2 g/mole = 0.318 grams H2 needed

Explanation:

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In a reaction, a + b + c→ d, it is found that the reaction is first order in terms of a and second order in terms of b and half

Vikentia [17]

Rate=[a]*([b]^2)*([c]^(1/2)]
rate=[2a]*([b]^2)*([2c]^(1/2)]= 2*(2^(1/2)[a]*([b]^2)*([c]
it increases times 2*(2^(1/2)=2√2

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2 years ago

What is name of first man made element?

enot [183]

The first element discovered through synthesis was technetium—its discovery being definitely confirmed in 1936. Hope that helps.

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2 years ago

Read 2 more answers

How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?

allochka39001 [22]

The complete balanced chemical reaction is:

2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S

First let us calculate the number of moles of AgNO3.

moles AgNO3 = 0.315 M * 0.035 L

moles AgNO3 = 0.011025 mol

From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

moles Na2S required = 5.5125 x 10^-3 mol

Therefore volume required is:

volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

volume Na2S = 0.0212 L = 21.2 mL

6 0

3 years ago

What is the pH of a 8.7 x 10^-12 M OH- solution ? pH= ?

vivado [14]

PH=14-pOH

pOH=-lg[OH⁻]

pH=14+lg[OH⁻]

pH=14+lg(8.7*10⁻¹²)=14-11.06=2,94

5 0

2 years ago

Read 2 more answers

A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base

yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

$pK_{b}=4.82$

$K_{b}=10^{-4.82}=1.5136\times 10^{-5}$

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.805\ g}{82.0343\ g/mol}$

$Moles= 0.022\ moles$

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.022}{0.055}$

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136\times 10^{-5}=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

pH = 14 - pOH = 14 - 2.61 = 11.39

5 0

2 years ago