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What is the mass of aluminum metal that reacts to give 11.1g of manganese metal?
Wiki User
∙ 14y ago
What is the mass of aluminum metal that reacts to give 11.1g of manganese metal?
Wiki User
∙ 14y ago
Jessica Apakama
7.27 g Al
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What is the gram formula mass of CaCl2?
111g
Molarity of a solution containing 0.456 grams CaCI in 250ml of solution?
First off we multiply .456 by 4 to get it into weight per litre. This gives us 1.824 grams per litre. Next, I'll assume you made a mistake here. Ca is 2+ so you need TWO chlorines (Cl-). I.e. the molecular weight is actually 111g/mol. 1.824/111 is 0.01643 molar. Be careful, because the wrong molecular weight can throw your answer off hugely, especially a heavy atom such as chlorine.
How do you calculate the mole of reactant and product?
Starting off with masses for each you use the mass-mole relationship n=m/M, where n is the number of moles of a substance (mol), m is the starting mass of the substance (g), and M is the MOLAR mass of the substance (g/mol). BALANCED REACTION 2NaOH(aq) + CaCl2(aq) ----> 2NaCl + Ca(OH)2 Case 1: Sodium hydroxide is the LIMITING reagent (its molar amount is less than twice the amount of calcium chloride), i.e. NaOH = 5.00 g and CaCl2 = 3.00 g In this case we use the mass of NaOH to find the number of moles. n=m/M=5.00g/40.0g/mol=0.125mol From here we compare molar ratios of the reaction (stoichiometry) to find what the corresponding number of moles of each product will be when the reaction ENDS (at equilibrium). NaCl:NaOH = 2:2 ratio = 1:1, therefore the number of moles of NaCl will be the same at the END of the reaction as the NaOH at the START of the reaction: 0.125 mol. Ca(OH)2:NaOH = 1:2, therefore there will be half as many moles since it takes two moles of reactant to create one mole of product (as dictated by the reaction above): 0.0625 mol Case 2: Calcium chloride is the LIMITING reagent (its molar amount is less than half the amount of sodium hydroxide), i.e. NaOH = 9.00 g and CaCl2 = 3.00 g n=m/M=3.00g/111g/mol=0.0270 mol Just as in Case 1, we compare molar ratios of the reaction (stoichiometry) to find what the corresponding number of moles of each product will be when the reaction ENDS (at equilibrium). NaCl:CaCl2 = 2:1 ratio, therefore there will be twice as many moles since it takes one mole of reactant to create two moles of product (as dictated by the reaction above): 0.0540 mol Ca(OH)2:CaCl2 = 1:1 ratio, therefore the number of moles of NaCl will be the same at the END of the reaction as the NaOH at the START of the reaction: 0.0270 mol. Keep in mind this only works with a BALANCED chemical reaction.
What is the gram formula mass of CaCl2?
111g
What is 100g plus 10g plus 1g?
100g + 10g + 1g = 111g
What is the relative formula mass for CaSO4?
136.1406 g/mol
A solution containing 14 grams of AgNo3 is added to a solution containing 4.83 g of CaCl2. Find the mass of the precipitate produced.?
Ok, lets begin by writing out the reaction : 2AgNO3 +CaCl2 --> 2AgCl(s) + Ca(NO3)2 Precipitate = AgCl Now find the mol of compound in each solution: 14g AgNO3 x (mol/170g) = .082mol 4.83g CaCl2 x (mol/111g) = .044mol Determine limiting reactant: Notice in reaction that 2 CaCl2 molecules react with 1 AgNO3. Because 2(.044mol) > 1(.082mol), AgNO3 is your limiting reactant. Now that you know this you can find the mass of the precipitate .082molAgNO3x (2molAgCl/2molAgNO3)x(143.3g/molAgCl) = 11.75g b) Assuming all the AgNO3 is exhausted, there will be 2(.044)-(.082) = .006mol CaCl2 left .006mol x (111g/mol) = 0.67g CaCl2
Molarity of a solution containing 0.456 grams CaCI in 250ml of solution?
First off we multiply .456 by 4 to get it into weight per litre. This gives us 1.824 grams per litre. Next, I'll assume you made a mistake here. Ca is 2+ so you need TWO chlorines (Cl-). I.e. the molecular weight is actually 111g/mol. 1.824/111 is 0.01643 molar. Be careful, because the wrong molecular weight can throw your answer off hugely, especially a heavy atom such as chlorine.
Which cars have the lowest carbon emissions?
1. Peugeot iOn - 88g/kmAs a fully electric car, nothing actually comes out of this vehicle while it is being driven, but CO2 is released through the production of the electricity used to charge it. We used the latest figures from the Department for Energy and Climate Change, which state that an average of 521 grams of CO2 are emitted per kilowatt hour of electricity, to calculate the iOn's CO2 emissions. 2. Volkswagen Polo - 99g/kmAll the VW Polos we've tested have low emissions, but it's the 1.2 TDI BlueMotion diesel that emits the least at 99g/km. This is exceptionally good and contributes to low running costs. In contrast, the 1.4-litre petrol engine is the worst Polo we've tested for emissions, recording 140g/km. 3. Vauxhall Ampera - 102g/kmThe Ampera is a 'range-extender' hybrid. This means is can run on electricity with no range anxiety, as it also has a petrol engine. Short distances can be completed in electric mode only, so there are zero emissions. However, according to our tests of the car in combined electric/petrol mode, it actually emits 102g/km of CO2. 4. Smart ForTwo - 105g/kmThe Smart ForTwo is one of the world's most recognisable city cars. It's only got two seats, but is surprisingly spacious for its driver and passenger. The 0.8 CDI engine promises low carbon dioxide emissions levels, and it recorded 105g/km in our tests. However, the official manufacturer CO2 claim is 88g/km, so this car still benefits from free road tax. 5. Nissan Leaf - 106g/kmOne of the first battery-driven, family-sized cars available in the UK, the Leaf offers low running costs. Using the same system as the Peugeot iOn, we measured the emissions at 106g/km. However, the precise CO2 output of the car will vary depending on exactly how it is charged. For example, if charging is supplemented by solar power, then that figure will be lower. 6. Volkswagen Golf - 108g/kmThe VW Golf has a wide range of emissions depending on which engine you opt for. The 1.4 TFSI recorded a lofty 160g/km. However, the 1.6TDI BlueMotion impressed our experts with only 108g/km CO2 emissions, which ensures car tax is kept to a minimum. 7. Toyota Auris - 109g/kmToyota launched this hybrid version of its Auris back in 2010. The Auris hybrid is smooth and quiet, and has so far proved very reliable. As for emissions, there is still some work to be done to hit the 2020 target, but at 109g/km it is one of the best performers at present. 8. Honda Insight - 111g/kmRivaling the Toyota Prius in the large cars class, the Insight hybrid has an efficient 1.3-litre engine. CO2 emission claims of the latest Insight are between 96g/km and 99g/km (bettering the original version's 101g/km), however our tests recorded 111g/km. =10. Audi A1 - 111g/kmTaking manufacturer claims, no Audi A1 has official C02 emissions of more than 125g/km, meaning you'll pay no more than £95 a year in tax. The most efficient engine we've tested is the 1.6 TDI - measuring 111g/km, 8g.km higher than the claim. =10. Toyota Prius Plug-In - 111g/kmThe brand new Prius Plug-In builds on the original Prius's eco-friendly reputation. The hybrid petrol-electric 1.8-litre engine offers electric-only driving at speeds up to 62mph. But, once the petrol kicks in, emissions increase to a recorded 111g/km.
What is the molarity of a solution that contains 9.63 grams of HCl in 1.5 liters of solutions?
Molarity = moles of solute/Liters of solution Find moles NaCl 9 grams NaCl (1 mole NaCl/58.44 grams) = 0.154 moles NaCl Molarity = 0.154 moles NaCl/1 Liter = 0.2 M sodium chloride -------------------------------
How do you calculate the mole of reactant and product?
Starting off with masses for each you use the mass-mole relationship n=m/M, where n is the number of moles of a substance (mol), m is the starting mass of the substance (g), and M is the MOLAR mass of the substance (g/mol). BALANCED REACTION 2NaOH(aq) + CaCl2(aq) ----> 2NaCl + Ca(OH)2 Case 1: Sodium hydroxide is the LIMITING reagent (its molar amount is less than twice the amount of calcium chloride), i.e. NaOH = 5.00 g and CaCl2 = 3.00 g In this case we use the mass of NaOH to find the number of moles. n=m/M=5.00g/40.0g/mol=0.125mol From here we compare molar ratios of the reaction (stoichiometry) to find what the corresponding number of moles of each product will be when the reaction ENDS (at equilibrium). NaCl:NaOH = 2:2 ratio = 1:1, therefore the number of moles of NaCl will be the same at the END of the reaction as the NaOH at the START of the reaction: 0.125 mol. Ca(OH)2:NaOH = 1:2, therefore there will be half as many moles since it takes two moles of reactant to create one mole of product (as dictated by the reaction above): 0.0625 mol Case 2: Calcium chloride is the LIMITING reagent (its molar amount is less than half the amount of sodium hydroxide), i.e. NaOH = 9.00 g and CaCl2 = 3.00 g n=m/M=3.00g/111g/mol=0.0270 mol Just as in Case 1, we compare molar ratios of the reaction (stoichiometry) to find what the corresponding number of moles of each product will be when the reaction ENDS (at equilibrium). NaCl:CaCl2 = 2:1 ratio, therefore there will be twice as many moles since it takes one mole of reactant to create two moles of product (as dictated by the reaction above): 0.0540 mol Ca(OH)2:CaCl2 = 1:1 ratio, therefore the number of moles of NaCl will be the same at the END of the reaction as the NaOH at the START of the reaction: 0.0270 mol. Keep in mind this only works with a BALANCED chemical reaction.
