a piece of metallic iron (10 moles) was dissolved in concentrated hydrochloric acid. the reaction formed iron (II) chloride and hydrogen gas. how many moles of iron (II) chloride were formed?
Fe + 2HCl ==> FeCl2 + H2
1 mol Fe = 1 mol FeCl2; therefore,
10 mol Fe = 10 mols FeCl2.
a piece of metallic iron (moles) was dissolved in concentrated HCL.The reaction form H2 and FeCI2 write and balance the equation for this reaction and determine the amount of formed FeCL2 and the amount of used HCI
To determine the number of moles of iron (II) chloride formed, we need to use the balanced chemical equation for the reaction between metallic iron and hydrochloric acid:
Fe + 2HCl → FeCl2 + H2
From the balanced equation, we can see that one mole of iron (Fe) reacts with two moles of hydrochloric acid (2HCl) to produce one mole of iron (II) chloride (FeCl2) and one mole of hydrogen gas (H2).
Given that we have 10 moles of metallic iron (Fe), we can calculate the number of moles of iron (II) chloride formed:
10 moles Fe × (1 mole FeCl2 / 1 mole Fe) = 10 moles FeCl2
Therefore, 10 moles of iron (II) chloride were formed.
To find out how many moles of iron (II) chloride were formed, we first need to understand the balanced equation for the reaction between metallic iron and hydrochloric acid.
The balanced equation for this reaction is:
Fe + 2HCl → FeCl2 + H2
From the balanced equation, we can see that 1 mole of metallic iron reacts with 2 moles of hydrochloric acid to produce 1 mole of iron (II) chloride and 1 mole of hydrogen gas.
Given that you have 10 moles of metallic iron, we can use this stoichiometry to determine the number of moles of iron (II) chloride formed.
Since the molar ratio between iron (II) chloride and metallic iron is 1:1, we can conclude that 10 moles of iron (II) chloride were formed in the reaction.