Chemistry, 03.07.2019 11:50 bella6786

Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. if 3.38 mol of magnesium and 3.38 mol of hydrochloric acid are reacted, how many grams of magnesium chloride are produced?

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Answer from: SkinnestXOXO

The mass of magnesium chloride produced in the given reaction is 16.09 grams.

Explanation:

We are given:

Moles of magnesium = 3.38 mol

Moles of hydrochloric acid = 3.38 mol

The chemical equation for the reaction of magnesium and hydrochloric acid follows:

$Mg+2HCl\rightarrow MgCl_2+H_2$

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of magnesium.

So, 3.38 moles of hydrochloric acid will react with = $\frac{1}{2}\times 3.38=1.69mol$ of magnesium.

As, given amount of magnesium is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

2 moles of hydrochloric acid is producing 1 mole of magnesium chloride.

So, 3.38 moles of hydrochloric acid will react with = $\frac{1}{2}\times 3.38=0.169moles$ of magnesium chloride.

Now, calculating the mass of magnesium chloride by using equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of magnesium chloride = 95.21 g/mol

Moles of magnesium chloride = 0.169 moles

Putting values in above equation, we get:

$0.169mol=\frac{\text{Mass of magnesium chloride}}{95.21g/mol}\\\\\text{Mass of magnesium chloride}=16.09g$

Hence, the mass of magnesium chloride produced in the given reaction is 16.09 grams.

Answer from: Quest

answer: enthalpy

explanation:

for a exothermic reaction, the energy is released i.e. the energy of the products is lower than the energy of the reactants.

$\delta h=e_p-e_r$

$\delta h$ = enthaply

$e_p$ = energy of products

$e_r$ = energy of reactants

as $e_p< e_r$ , thus $\delta h$ comes out to be negative. negative value of enthalpy shows that the reaction is exothermic.

Answer from: Quest

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

= 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

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