Balance Al2O3 ----->Al+O2?
For the best answers, search on this site https://shorturl.im/a4G4Q 2 Al2O3 → 4 Al + 3 O2 (100 g Al) / (26.98154 g Al/mol) x (2 mol Al2O3 / 4 mol Al) x (101.96137 g Al2O3/mol) = 189 g Al2O3
How do you balance Al2O3+HCl=AlCl3+H2O?
First, look at Al. You have 2 on the left but only 1 on the right, so put a 2 in front of AlCl3. Because of that, you now have 6 Cl atoms on the right, but only 1 on the left, so put a 6 in front of HCl. Now you have 6 H atoms on the left but only 2 on the right, so put a 3 in front of H2O. Finally, you will notice that there are 3 O atoms on the right and 3 O atoms on the left, so you are done. Double check by counting how many of each atom there are on each side to follow the Law of the Conservation of Matter.
How do I balance Al2O3 + HCL ------> AlCl3 + H2O ?
count huge style the style of aspects on the two aspects of the equation. Use coefficients to regulate. verify you multiply subscripts and coefficients. Pb(NO3)2-----> PbO + NO2 + O2 Pb a million - a million N 2 - a million O 6 - 5 considering O has an peculiar huge style on the main appropriate, multiply PbO via 2, then stability Pb on the left area. 2 Pb(NO3)2-----> 2 PbO + NO2 + O2 Pb 2 - 2 N 4 - a million O 12 - 6 next stability N via including 4 in front of NO2. This additionally balances O. 2 Pb(NO3)2-----> 2 PbO + 4 NO2 + O2 Pb 2 - 2 N 4 - 4 O 12 - 12 you're able to do something. it is your HW and that's easy, applicable?
First take a look at the equationDetermine, before balancing, that all the formulas are written correctly. In this case the formulas are written correctly.The equation as it is now stands has one chlorine atom and one hydrogen atom on the left. There are two atoms of hydrogen and chlorine on the right.Now we must follow the rule:DO NOT CHANGE THE FORMULASDON’T CHANGE THE SUBSCRIPTTo balance the equation, place whole numbers ahead of the formulas so as to have the same number of all elements on each side.For example:In our original equation we haveMg + 2 HCl → MgCl2 + H2We place the number 2 ahead of the HCl. We now have 2 Hydrogens on each side and 2 chlorines on each side. The is also one Magnesium on each side. The equation is balanced.
Balance equation:Al+HCl=AlCl3+H2?
There are 3 Cl on the right, that's the highest number, so I'd target that first. x(Al) + 3(HCl) = AlCl3 + 1.5(H2) But that would give us 3 H's on the right, and we want H to be a multiple of 2. So double the HCl again. x(Al) + 6(HCl) = 2(AlCl3) + 3(H2) At this point, you can see how many Al you need on the left.
Can you balance these equations?
Answers first: 1) PbS + 2PbO -----> 3Pb + SO2 2) Al2O3 + 6HCl ----> 2AlCl3 + 3H2O 3) 2Bi(NO3)3 + 3H2S ----> Bi2S3 + 6HNO3 I always try to think of an equation like a see-saw (not too condescending I hope!). Working through equation 1... PbS + PbO ---> Pb + SO2 I look at the right and see what I have. that will tell you what you need on the left. your first instinct is to say "I need another oxygen on the left to make both sides balance". to do that, simply put a 2 infront of the PbO. REMEMBER you can only change MOLAR RATIOS (i.e. large numbers infront of molecules) NOT numbers of bonded atoms (i.e. little numbers after the atom). That would change bonding etc and you don't want to mess with that. Enough rambling! So: PbS + 2PbO ----> Pb + SO2 great! We've balanced the oxygen's like we wanted, but now we have 3 Pb atoms on the left and only 1 on the right. So, it makes sense to write the number 3 infront of the Pb on the right to give: PbS + 2PbO --> 3Pb + SO2 nice and simple. All atoms on the left have the same number of atoms on the right. balanced see-saw. Equation 2 only appears a bit more tricky because there are more atoms involved. here we go again... Al2O3 + HCl --> AlCl3 + H2O need three oxygens on the right (as is on the left) need three chlorines on the left (as is on the right). Al2O3 + 3HCl --> AlCl3 + 3H2O uh-oh. the hydrogens are unbalanced (3 on the left; 6 on the right). Try doubling the value on the left. Al2O3 + 6HCl --> AlCl3 +3H2O chlorines are unbalanced (6 left; 3 right). stick a 2 infront of AlCl3 Al2O3 + 6HCl --> 2AlCl3 + 3H2O Piece of cake! fancy giving the last one a go yourself..? I'm tired. Hope that helps! Chemistry rules :)
Aluminum reacts with hydrochloric acid gives aluminum tri chloride and hydrogen gas. The balanced equation for this reaction is2Al + 6HCl --> 2AlCl3 + 3H2Balancing Equations Calculator
4 AL + 3O2 -> 2 Al2O3
Al(OH)3+HCl=AlCl3+H2O?
Step 1 balance equation Al(OH)3 + 3HCl = AlCl3 + 3H2O .750/ 78.0037(FM) .0096149285 moles of Al(OH)3 .0096149285 x mole ratio (unknown over known) so in this case the mole ratio of HCl is 3/1 moles of HCl=.00288447856 moles to grams= x FM .00288447856 x 36.4611= 1.05g of HCl .750/ 78.0037(FM) .0096149285 moles of Al(OH)3 .0096149285 x mole ratio (unknown over known) so in this case the mole ratio of H20 is 3/1 moles of H20=.00288447856 moles to grams= x FM .00288447856 x 18= .0519g of H20
How do I balance this chemical equation? Al + HCl -> AlCl3 + H2?
Unbalanced equation: Al + HCl → AlCl3 + H2 To balance things, you have to have the same amount of atoms on each side. So ask yourself: How many Al atoms do I have on both sides? 1 on the reactant side and 1 on the product side. Check. How many Cl atoms do I have on both sides? 1 on the reactant side and 3 on the product side. I need to have 3 on both sides, so I have to multiply HCl by 3. Al + 3 HCl → AlCl3 + H2 How many H atoms do I now have on both sides? 3 on the reactant side and 2 on the product side. I need to have the same amount on both sides, so I have to double everything. 2 Al + 6 HCl → 2 AlCl3 + H2 Okay, NOW how many H atoms do I have? 6 on reactant side, 2 on product side. So I can just multiply H2 by 3 to balance everything. 2 Al + 6 HCl → 2 AlCl3 + 3 H2 How many moles of H2 are produced by the complete reaction in 186g of Al? 2 Al + 6 HCl → 2 AlCl3 + 3 H2 You have 186 g of Al. The molar mass of Al is 26.98 grams/mole. That means you have (186 g)/(26.98 g/mol) = 6.89 mol of Al. As per the balanced equation, Al and H2 react in a 2:3 ratio. So you would have 6.89 mol × (3/2) ≈ 10.3 mol of H2.