$Mg$ 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?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

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 moles of $"Mg"$.

$1.016 cancel("mol O"_2) × "2 mol Mg"/(1 cancel("mol O"_2)) = "2.031 mol Mg"$

Step 3. Calculate the mass of $"Mg"$.

$2.031 cancel("mol Mg") × ("24.30 g Mg")/(1 cancel("mol Mg")) = "49.4 g Mg"$

The reaction required $"49.4 g Mg"$.

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
Aluminum will react with sulfuric acid in the following reaction: $2Al(s) + 3H_2SO_4(l) -> Al_2(SO_4)_3(aq) + 3H_2(g)$. How many moles of $H_2SO_4$ will react with 18 mol Al? How many moles of each product will be produced?

With reference to the given balanced equation for the , the needed number of moles for each remaining compound involved in the reaction using the mass of $Al$ as basis...

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
In the reaction $2Mg + O_2 -> 2MgO$, the law of definite proportions states that for every 2 moles of $Mg$ you will need how many moles of $O_2$?

It's in the balanced equation. Each number before a compound corresponds to how many miles you use, and when there is no number you put in a 1. So 2...

1 Answers 1 views
How many moles of magnesium oxide are formed when 4 moles of magnesium react with oxygen in the reaction $2Mg + O_2 -> 2MgO$?

Let us first look at the equation. As per the equation , 2 moles of Magnesium produces two moles of MgO . If we double the amount of Magnesium ,...

1 Answers 1 views
For the reaction,$2x$+$3y$+$4z$$ rarr$$ 5w$ Initially if 1 mole of $x$,3 mole of $y$ and 4 mole of $z$ is taken and 1.25 mole of $w$ is obtained then what is the % yield of this reaction?

This is a limiting reactant problem. We know that we will need a balanced equation and moles of each reactant. 1. Gather all the information in one place with...

1 Answers 1 views
When a container is filled with 3.00 moles of $H_2$, 2.00 moles of $O_2$, and 1.00 mole of $N_2$, the pressure in thecontainer is 768 kPa. What is the partial pressure of $O_2$?

Partial Pressure of $O_2$ = 256 kPa. Dalton's law of partial pressures covers two concepts; Mole fraction Partial pressure Let's talk about Mole Fraction first. You...

1 Answers 1 views
A sample containing 4.80 g of $O_2$ gas has a volume of 15.0 L at constant pressure and temperature. What is the new volume, in liters, after 0.500 mole of $O_2$ gas is added to the initial sample of 4.80 g of $O_2$?

Initially, you have $4.80color(red)(cancel(color(black)("g O"_2))) × "1 mol O"_2/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.1500 mol O"_2$ Thus, you have $"0.1500 mol O"_2/"15.0 L" = "0.0100 mol O"_2/"1 L"$ If you...

1 Answers 1 views
When 2 moles of $H_2(g)$ and 1 mole of $O_2(g)$ react to give liquid water, 572 kJ of heat evolve. $2H_2(g)$ + $O_2(g)$ $rarr$ $2H_2O(l)$ ; $DeltaH$ = -572 kJ a) What is the themochemical eq. for 1 mole of liquid water?

You know that when $2$ moles of hydrogen gas react with $1$ mole of oxygen gas, you get $2$ moles of water and $"572 kJ"$ of heat are evolved....

1 Answers 1 views
X