I have a Chemistry question please help me for 10 points and 5 stars?

1. How many moles of hydrogen gas are needed to react with 12.1 g chlorine gas, too produce hydrogen chloride gas?





Equation

I have a Chemistry question please help me for 10 points and 5 stars?
Cl2(g) + H2(g) ---%26gt; 2HCl(g)





So is a 1: 1 ratio of Chlorine reacting with Hydrogen.





Mr of Cl2 = 2 x 35.5 = 71


Mr of H2 = 2 x 1 = 2





Moles of Cl2 = mass/Mr


= 12.1 g / 71


= 0.1704 moles





Mass of hydrogen = moles x Mr


= 0.1704 moles x 2


= 0.3408 moles


0.34 moles would do for 2d.p answer





Hope this helps :o)
Reply:the equation for this reaction is:





H2+Cl2 -%26gt; 2HCl





mass of chlorine=12.1g


molar mass of hydrogen=2 since its a di atom


molar mass of chlorine=35.5+35.5=71g/mol (since chlorine is a di atom)


number of moles=mass/molar mass


number of moles of chlorine=12.1/71


number of moles of chlorine=0.17mol





from mole ratio in the balanced equation


number of moles of chlorine: number of moles of hydrogen


= 1:1(from the balanced equation)


therefore 0.17 mol of Chlorine reacted with 0.17 mol of Hydrogen





0.17 mols of hydrogen are needed to react with 12.1g of chlorine gas





If u want to know da mass of Hydrogen thereafter, simply multiply its moles by the molar mass i.e


mass=molesxmolar mass


mass=0.17x2


mass=0.34g


but you asked for only moles so i hope i've done my best!!!
Reply:Alright, fair enough...





This is a composition reaction where Chlorine and Hydrogen combine to produce Hydrogen Chloride gas, so we need to start with the equation. Noting that both Chlorine and Hydrogen are diatomic, we arrive with the following formula:





Cl2 + H2 =====%26gt; 2HCl





That is, 1 mole of Chlorine gas combining with 1 mole of Hydrogen gas will yield 2 moles of Hydrogen Chlorine gas. (Coefficients represent mole amounts).





The question gives us the amount of chlorine gas in grams, and asks for the number of moles of hydrogen... so this sets up our stoichiometry problem:





12.1g Cl2 x (1 mol Cl2 / 35.45g Cl2) x (1 mol H2 / 1 mol Cl2)





by using the factor-label method and using the mole ratio of 1:1, we arrive at .34 mol H2





Hope this helps.

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