For the decomposition reaction $2Ag_2CO_3->4Ag+2CO_2+O_2$, how many moles of reactant undergo decomposition in order to produce 6.0 moles of $Ag$?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

The reaction above correctly represents thermal decomposition of a metal carbonate. The carbonate gives metal oxide and carbon dioxide under STRONG heating. You are not going to get silver metal unless you get it all into solution and add some reductant (maybe copper). Of course, you could do this in a forge, but it would be very difficult to smelt; and you'd still have to add a reductant.

Instead, we could make silver nitrate:

$Ag_2O + 2HNO_3(aq) rarr 2AgNO_3(aq) + H_2O(l)$

Silver nitrate would be relatively easy to reduce.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
The reaction between $H_2$ and $O_2$ produces 13.1 g of water. How many grams of $O_2$ reacted? $2H_2(g) + O_2(g) -> 2H_2O(g)$?

There are $(13.1*g)/(18.01*g*mol^-1)=0.727*mol$ of water product. $H_2(g) + 1/2O_2(g) rarr H_2O(l)$ Given the of the reaction, there were necessarily $0.727*mol$ dihydrogen reactant, and $(0.727*mol)/2$ dioxygen reactant. I speak of dihydrogen,...

1 Answers 1 views
How many grams of $Ag$ can be formed from 5.50 grams of $Ag_2O$ in the equation: $2Ag_2O (s) -> 4Ag (s) + O_2 (g)$?

$"Moles of silver oxide"$ $=$ $(5.50*g)/(231.74*g*mol^-1)$ $=$ $??mol$ $"Grams of silver metal"$ $=$ $2xx(5.50*g)/(231.74*g*mol^-1)xx107.87*g*mol^-1$ $=$ $??*g$. Why did I put the $2$ in the front of the expression?

1 Answers 1 views
$2CO(g) + O_2(g) -> 2CO_2(g)$. How many liters of $O_2$ will react with 74.2 liters of CO? If 6.5 g of CO react, how many liters of $CO_2$ are produced?

We calculate the relative quantities from the molar ratios in the balanced reaction equation. $2CO(g)+O_2(g)→2CO_2(g)$. As a pretty good approximation, gas volumes are proportional to molar amounts. Therefore with...

1 Answers 1 views
How many moles of $CO$ and $O_2$ are required to produce 10 moles of $CO_2$ in the reaction $2CO + O_2 -> 2CO_2$?

Balanced equation $"2CO"("g") + "O"_2("g")"$$rarr$$"2CO"_2("g")"$ To calculate moles $"CO"$ and $"O"_2"$ required to produce $"10 mol CO"_2"$, multiply mol $"CO"_2"$ by the mol ratio between the desired reactant and $"CO"_2"$...

1 Answers 1 views
$Mg$ and $O_2$ react in a 2.1 molar ratio. 2 moles $M_g$ = 1 mole $O_2$. If a reaction used 32.5 g of $O_2$, how many g of $Mg$ reacted?

Step 1. Calculate the moles of $"O"_2$. $32.5 cancel("g O"_2) × ("1 mol O"2)/(32.00 cancel("g O"2)) = "1.016 mol O"_2$ Step 2. Use the molar ratio to calculate the...

1 Answers 1 views
Given the equation $2H_2O ->2H_2 + O_2$, how many moles of $H_2O$ would be required to produce 2.5 moles of $O_2$?

Since the chemical equation, $color(red)2H_2O->2H_2+color(blue)1O_2$ requires a specific number of moles of the reactant, then a specific number of moles of the products are created. With this knowledge,...

1 Answers 1 views
Given the equation $2H_2O -> 2H_2 + O_2$, how many moles of $H_2O$ be required to produce 2.5 moles of $O_2$?

So every 2 molecules of water gives 1 molecule of oxygen. This ratio ($2:1$) stays the same if we are talking about moles, because a mole is a set number...

1 Answers 1 views
Given the balanced equation representing a reaction: $2CO(g) + O_2(g) -> 2CO_2(g)$. What is the mole ratio of $CO(g)$ to $CO_2(g)$ in this reaction?

We could write...$CO(g) +1/2O_2 rarr CO_2(g)$, instead of ... $2CO(g) +O_2 rarr 2CO_2(g)$ In either scenario both mass and charge are balanced, as required for a stoichiometric equation...and this is...

1 Answers 1 views
In the reaction $2CO(g) + O_2(g) -> 2CO_2(g)$, what is the ratio of moles of oxygen used to moles of $CO_2$ produced?

You have given the stoichiometric equation, which states UNEQUIVOCALLY that 2 mole of carbon monoxide reacts with 1 mole of dioxygen gas to give 2 mole of carbon dioxide. We...

1 Answers 1 views
$2H_2O -> 2H_2 + O_2$ What is the percent yield of $O_2$ if 10.2 g of $O_2$ is produced from the decomposition of 17.0 g of $H_2O$?

You know by looking at the balanced chemical equation $2"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))$ that it takes $2$ moles of water to produce...

1 Answers 1 views
X