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1 mole of sulfuric acid=98g(?)chemical formula of sulfuric acid=H2SO4Atomic Mass of Hydrogen(H)=1
Atomic mass of Sulpher(S)=32
Atomic mass of Oxygen(O)=16
(1)2+32+(16)4
=2+32+64
=98u
To calculate number of moles
No. of moles=given mass/atomic mass
no. of moles=49/98
no. of moles is 0.5g
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What is the mass of water when 4gm hydrogen and 64gm oxygen combined together nd how?
Balanced equation first.2H2 + O2 --> 2H2OGet moles products.4 grams H2 (1 mole H2/2.016 grams) = 1.984 moles H264 grams O2 (1 mole O2/32 grams) = 2.000 moles O2I suspect hydrogen gas of limiting and driving the reaction.1.984 moles H2 (1 mole O2/2 moles H2) = 0.992 moles O2 ( you have more than this in equation )2.000 moles O2 (2 mole H2/1 mole O2) = 4.000 moles H2 ( you do not have this much and H2 will drive this reaction )1.984 moles H2 (2 moles H2O/2 moles H2)(18.016 grams/1 mole H2O)= 36 grams water produced====================
How many grams of H2 are needed to produce 10.06 g of NH3?
The formula reaction for NH3 when using N2 and H2 is: (N2)+3(H2) ---> 2(NH3) Now, first step is to find the moles of the H2 reactant. This is found via (grams of reactant)/(molar mass of reactant). There are 10 grams, and the molar mass of H2 is approximately 2.016. Therefore, the equation should look like: 10/2.016. This yields a value of ~4.9606 moles of H2. Now, you use the molar ratio from the reactant to the product to determine how many moles of product were yielded. According to the reaction, three moles of H2 are required to produces 2 moles of NH3. So, the mole ratio is 2/3. Multiply the number of moles of H2 with the molar ratio to determine the moles of NH3. 4.0606 * 2/3 = 3.3071 moles of NH3. Multiply the number of moles with the molar mass of NH3 (17.0306), and voila! 3.3071 * 17.0306 = 56.3216 grams. Now, if your teacher is feeling like a stickler about significant figures, than that value should be rounded to 56 grams of NH3.
How many moles of H2O are product from reaction of 355 g of O2?
Balanced equation. 2H2 + O2 -> 2H2O 355 grams O2/32 grams = 11.1 moles O2 check for limiting reactant 11.1 moles O2 (2 mole H2/1 mole O2) = 22.2 mole H2 and H2 has no where near that many moles, so limits and drives reaction so, as they are one to one...... 22.2 moles of H2O are produced
How many grams of H2 are needed to reduce 5.51 moles of CuO in the reaction?
11 g hydrogen are needed.
How many moles of nh3 are produced from 1.5 moles h2?
1 mole
How many grams of hydrogen are needed to produce 30.0 grams of water?
The synthesis reaction is 2 H2 + O2 = 2 H2O. Every two moles of hydrogen reacts with one mole of oxygen to make two moles of water. Then 30.0 grams of water is 1.67 moles, and 1.67 moles of H2 has a mass of 3.37 grams. 25.0 grams of O2 is .781 moles, so 1.562 moles of H2 are needed, or 3.15 grams.
What is the mass of water when 4gm hydrogen and 64gm oxygen combined together nd how?
Balanced equation first.2H2 + O2 --> 2H2OGet moles products.4 grams H2 (1 mole H2/2.016 grams) = 1.984 moles H264 grams O2 (1 mole O2/32 grams) = 2.000 moles O2I suspect hydrogen gas of limiting and driving the reaction.1.984 moles H2 (1 mole O2/2 moles H2) = 0.992 moles O2 ( you have more than this in equation )2.000 moles O2 (2 mole H2/1 mole O2) = 4.000 moles H2 ( you do not have this much and H2 will drive this reaction )1.984 moles H2 (2 moles H2O/2 moles H2)(18.016 grams/1 mole H2O)= 36 grams water produced====================
How many moles of H2 are in 30 grams of Na?
None. There is no hydrogen in sodium
How many grams of H2 are needed to produce 10.06 g of NH3?
The formula reaction for NH3 when using N2 and H2 is: (N2)+3(H2) ---> 2(NH3) Now, first step is to find the moles of the H2 reactant. This is found via (grams of reactant)/(molar mass of reactant). There are 10 grams, and the molar mass of H2 is approximately 2.016. Therefore, the equation should look like: 10/2.016. This yields a value of ~4.9606 moles of H2. Now, you use the molar ratio from the reactant to the product to determine how many moles of product were yielded. According to the reaction, three moles of H2 are required to produces 2 moles of NH3. So, the mole ratio is 2/3. Multiply the number of moles of H2 with the molar ratio to determine the moles of NH3. 4.0606 * 2/3 = 3.3071 moles of NH3. Multiply the number of moles with the molar mass of NH3 (17.0306), and voila! 3.3071 * 17.0306 = 56.3216 grams. Now, if your teacher is feeling like a stickler about significant figures, than that value should be rounded to 56 grams of NH3.
How many grams of water can be prepared from 5moles of hydrogen at standard conditions?
Balanced equation. 2H2 + O2 --> 2H2O 5 moles H2 (2 moles H2O/2 mole H2)(18.016 grams/1 mole H2O) = 90 grams of water ===============
How many hydrogen atoms are present in 3.65 grams of water?
3.65 grams of water is equal to .203 moles of H2O. This means there is also .203 moles of H2 present, or .408 grams.
What is the percent yield of h2o if 58g h2o are produced by combining 60g o2 and 7g h2?
Balanced equation. 2H2 + O2 >> 2H2O ( now find limiting reactant ) 7 grams H2 (1 mole H2/2.016 grams) = 3.472 moles H2 60 grams O2 (1 mole O2/32 grams) = 1.875 moles O2 1.875 moles O2 (2 mole H2/1 mole O2) = 3.75 mole H2 ( checked O2, but I know H2 limits because you do not have 3.75 moles H2, so H2 drives reaction) 3.472 moles H2 (2 mole H2O/2 mole H2)(18.016 grams/1 mole H2O) = 62.552 grams H2O produced, so; 58 grams/62.552 times 100 = 92.7% yield of H2O, call it 93%
How many moles of H2O are product from reaction of 355 g of O2?
Balanced equation. 2H2 + O2 -> 2H2O 355 grams O2/32 grams = 11.1 moles O2 check for limiting reactant 11.1 moles O2 (2 mole H2/1 mole O2) = 22.2 mole H2 and H2 has no where near that many moles, so limits and drives reaction so, as they are one to one...... 22.2 moles of H2O are produced
How many grams of H2 are needed to reduce 5.51 moles of CuO in the reaction?
11 g hydrogen are needed.
How many moles of H2 are needed to produce 8.25 mole of H2O?
2H2 + O2 --> 2H2OFor every 2 moles of H2, 2 moles of H2O will be produced (i.e., a 1:1 ratio). So to produce 8.25 moles of H2O you will also need 8.25 moles of H2
How many moles of NH3 are produced when 1.4 moles H2 reacts?
Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
How many moles of hydrogen gas will be produced if 25g of Na reacts completely with excess HCl?
Balanced equation first. 2Na + 2HCl >> 2NaCl + H2 25 grams Na (1mol Na/22.99g )(1mol H2/2mol Na ) = 0.54 moles of H2 produced.
