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3 years ago

How many grams of nh3 can be produced from 3.80 mol of n2 and excess h2?

1 answer:

4 0

The balanced equation :N2 + 3H2 --> 2NH3

1) Let's find the molar mass of NH3 because it will be needed later.
To determine the molar mass you have to consult the periodic table and see the atomic mass of each element. If the element is repeted, for example 3 times, you multiple the atomic mass per 3.
NH3= 14 + 1 x 3= 17 g/mol

2) By looking at the balanced equation, you can see that 2 mol of NH₃ were produced with only one mol of N2. How much NH3 will result when we have 3,80 mol of N2?
So now we solve it with a porportion:

2 mol NH3-------1 mol of N2
x --------3.80 of N2

x= 2 x3,8 = 7,6mol NH3

3) the only thing left to do is convert the moles obtained, to grams. We can use the formula:
nº of moles x molar mass= mass
7,6mol NH3 x 17 g/mol= mass of NH3
mass= 129,2 grams of NH3

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A solution of sodium hydroxide was titrated against a solution of sulfuric acid. How many moles of sodium hydroxide would react

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2 mole of Sodium hydroxide reacts with 1 mole of Sulfuric acid

Explanation:

Write down the equation in the beginning with reactants and products:

NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Now try to balance it. Try with Na first:

2NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Na atoms are balanced. There are 6 Oxygen atoms on the right and 5 on the left. Balance by increasing the H₂O moles:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

Check if H atoms are also balanced. They are. That means our final reaction is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

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3 years ago

Consider the following reaction between mercury(II) chloride and oxalate ion.

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Answer: The rate law of the reaction is $\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_4^{2-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 2 from 1, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2$

Dividing 2 from 3, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2$

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Answer:

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