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2 years ago

2.20 moles Sn to grams

Chemistry

1 answer:

mihalych1998 [28]2 years ago

3 0

261.162 grams. Use the equation n=M÷Mr.

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Write a covalent compound formed by nitrogen and oxygen. what is its Lewis structure

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If 1. 3618 moles of AsF3 are allowed to react with 1. 0000 mole of C2Cl6, what would be the theoretical yield of AsCl3, in moles

VARVARA [1.3K]

Answer:

AsF3:C2CI6

4:3

1.3618 moles: 1.02135 moles(1.3618÷4×3)

C2CI6 is the limting reagent

So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)

Balanced equation

4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4

Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.

Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.

Explanation:

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2 years ago

if 0.40 mol of h2 and .15 mol of o2 were to reat as completely as possible to produce h20, what mass of the reactant would remai

Sveta_85 [38]

Answer:

0.2g

Explanation:

Given parameters:

Number of moles of H₂ = 0.4mol

Number of moles of O₂ = 0.15mol

Unknown:

Mass of reactant that would remain = ?

Solution:

To solve this problem, we need to know the limiting reactant which is the one in short supply in the given reaction.

The expression of the reaction is :

2H₂ + O₂ → 2H₂O

2 mole of H₂ will combine with 1 mole of O₂

But given; 0.4 mole of H₂ we will require $\frac{0.4}{2}$ = 0.2mole of O₂

The given number of oxygen gas is 0.15mole and it is the limiting reactant.

Hydrogen gas is in excess;

1 mole of oxygen gas will combine with 2 mole of hydrogen gas

0.15 mole of oxygen gas will require 0.15 x 2 = 0.3mole of hydrogen gas

Now, the excess mole of hydrogen gas = 0.4 mole - 0.3 mole = 0.1mole

Mass of hydrogen gas = number of mole x molar mass

Molar mass of hydrogen gas = 2(1) = 2g/mol

Mass of hydrogen gas = 0.1 x 2 = 0.2g

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2 years ago