Impossible conversion
1. Write the balanced equation: 2Al(s) + 6HCl(aq) ==> 2AlCl3(aq) + 3H2(g)2. Convert 3.70 g Al to moles of Al: 3.70 g x 1 mol/26.98 g 3. Use stoichiometric ratios in balanced equation to find moles H2 produced: answer from step 2 (moles Al) x 3 moles H2/2 moles Al = moles H2 produced 4. Convert moles H2 produced found in step 3 to grams H2 using molar mass of H2
H2 +Cl2---------------->2HCl Since H2 and Cl2 react in 1:1 mole ratio the number of moles of H2 reacting is equal to the number of moles of Cl2 which is equal to 0.213
First we are going to find number of H2 moles form it mass H2 moles=(2.37*10^-4)/2 = 1.185*10^-4moles Since 3 moles of H2 makes 2 moles of NH3 then by using this ration, we can find number of moles in NH3 NH3 moles= (2 * 1.185*10^-4)/3 = 0.79*10^-4 moles Finally, we find number of NH3 molecules by multiplying the number of moles with (6.022*10^23). #NH3 molecules = (0.79*10^-4) *(6.022*10^23) =4.75 * 10^19 molecules of NH3 Good luck Enas
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.
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====================
2 moles.
1. Write the balanced equation: 2Al(s) + 6HCl(aq) ==> 2AlCl3(aq) + 3H2(g)2. Convert 3.70 g Al to moles of Al: 3.70 g x 1 mol/26.98 g 3. Use stoichiometric ratios in balanced equation to find moles H2 produced: answer from step 2 (moles Al) x 3 moles H2/2 moles Al = moles H2 produced 4. Convert moles H2 produced found in step 3 to grams H2 using molar mass of H2
2.4
H2 +Cl2---------------->2HCl Since H2 and Cl2 react in 1:1 mole ratio the number of moles of H2 reacting is equal to the number of moles of Cl2 which is equal to 0.213
Given/Known:1mole of H2 = 2.01588g H21mole of H2 = 6.022 x 1023 molecules H21) Convert molecules of H2 to moles of H2 by doing the following calculation.9.4 x 1025 molecules H2 x (1mol H2/6.022 x 1023 molecules H2) = 156mol H22) Convert the moles of H2 to mass in grams of H2.156mol H2 x (2.01588g H2/1mol H2) = 314g H2
2,25 H2 moles
First we are going to find number of H2 moles form it mass H2 moles=(2.37*10^-4)/2 = 1.185*10^-4moles Since 3 moles of H2 makes 2 moles of NH3 then by using this ration, we can find number of moles in NH3 NH3 moles= (2 * 1.185*10^-4)/3 = 0.79*10^-4 moles Finally, we find number of NH3 molecules by multiplying the number of moles with (6.022*10^23). #NH3 molecules = (0.79*10^-4) *(6.022*10^23) =4.75 * 10^19 molecules of NH3 Good luck Enas
depends on how much aluminum oxide you have, 1gram, 2 billion Kg? how many? cant find the number of moles of oxygen without knowing the mass of the al203
Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
0.175 X Avogadro's Number = about 1.05 X 1023.
The reaction between nitrogen and hydrogen to form ammonia is: N2 + 3 H2 → 2 NH3 The above is the reaction for the Haber process in the industrial synthesis of ammonia. For a given proportion of 3 N2 to 2 H2 (or in ratio terms equivalent to 4.5 N2 to 3 H2), we see that H2 is the limiting reactant. Thus according to the stoichiometry of the reaction, 2 moles of H2 will form 1.33 moles 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.