Answer:- 448 mL of hydrogen gas are formed.
Solution:- It asks to calculate the volume of hydrogen gas formed in milliliters at STP when 0.020 moles of magnesium reacts with excess HCl acid. The balanced equation is:
There is 1:1 mol ratio between Mg and hydrogen gas. So, the moles of hydrogen gas is also equals to the moles of Mg reacted.
moles of Hydrogen gas formed = 0.020 mol
At STP, volume of 1 mol of the gas is 22.4 L. We need to calculate the volume of 0.02 moles of hydrogen gas.
= 0.448 L
They want answer in mL. So, let's convert L to mL using the conversion formula, 1L = 1000mL
= 448 mL
So, 0.020 moles of magnesium would produce 448 mL of hydrogen gas at STP on reacting with excess of HCl acid.
Answer:
4.43 g
Explanation:
The reaction between sodium chloride and flourine gas is given as;
NaCl + F2 --> NaF + Cl2
From the stochiometry of the equation;
1 mol of NaCl reacts eith 1 mol of F2 to form 1 mol of NaF and Cl2
Mass of 1 mol of F2 = 38g
Mass of 1 mol of sodium flouride, NaF = 42g
This means 38g of flourine reacted with NaCl to form 42g of NaF
xg of F2 would form 4.9g of NaF
38 = 42
x = 4.9
x = 4.9 * 38 / 42
x = 4.43 g
Answer:
Law of Conservation of Energy
Explanation:
Answer:
Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O
Explanation:
We have the products of a reaction and we have to predict the reactants. Since the products are binary salt and water, this must be a neutralization reaction. In neutralizations, acids react with bases. The acid that gives place to Br⁻ is HBr, while the base the gives place to Ba²⁺ is Ba(OH)₂. The balanced chemical equation is:
Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O
