An object that sticks to iron

If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for

Luba_88 [7]

The number of grams of Ag2SO4 that could be formed is 31.8 grams

<u><em> calculation</em></u>

Balanced equation is as below

2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)

Find the moles of each reactant by use of mole= mass/molar mass formula

that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles

moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles

use the mole ratio to determine the moles of Ag2SO4

that is;

the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles

The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles

AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles

<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>

that is 0.102 moles x 311.8 g/mol= 31.8 grams

3 0

3 years ago

Na2CO3 + 2HCl ---------> 2NaCl + CO2 + H2O How many moles of NaCl are produced from the reaction of 1.67 x 1022 molecules of

dexar [7]

Answer:

0.0554 moles of NaCl are produced from the reaction of 1.67*10²² molecules of Na₂CO₃ with excess HCl.

Explanation:

The balanced reaction is:

Na₂CO₃ + 2 HCl → 2 NaCl + CO₂ + H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

Na₂CO₃: 1 mole
HCl: 2 moles
NaCl: 2 moles
CO₂: 1 mole
H₂O: 1 mole

On the other hand, Avogadro's Number is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number applies to any substance.

In this case, you can apply the following rule of three: if 6.023*10²³ molecules of Na₂CO₃ are contained in 1 mole, 1.67*10²² molecules will be contained in how many moles?

$amount of moles=\frac{1.67*10^{22}molecules*1mole }{6.023*10^{23}molecules}$

amount of moles= 0.0277 moles

In this case, you can apply the following rule of three: if by stoichiometry 1 mole of Na₂CO₃ produces 2 moles of NaCl, 0.0277 moles of Na₂CO₃ will produce how many moles of NaCl?

$amount of moles of NaCl=\frac{0.0277 moles of Na_{2} CO_{3}*2 moles of NaCl }{1 mole of Na_{2} CO_{3}}$