For the reaction H2 + Cl2 = 2HCl how many moles HCl will be produced from...
How many liters of HCl(g) measured at STP can be produced from 4.00 g of Cl2 and excess H2 according to the following equation: H2(g) + Cl2(g) 2HCl(g)
Complete and balance the reaction below: Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (s) How many moles of H2 gas will be produced from 2 mol of HCl ? 2HCl ---à H2 + Cl2 1 mol of H2 gas will be produced from 2 moles HCl. How would you determine the volume of a 125mL Erlenmeyer flask that you will use for the experiment? (a 125 mL flask does not have a total volume of 125 mL)...
How many grams of HCl are produced according to the following equation, H2 + Cl2 → 2 HCl, when 4.0 g of hydrogen reacts completely? with work and explanations, please using conversion factors
For the reaction, calculate how many moles of the product form when 1.93 mol of H2 completely reacts. Assume that there is more than enough of the other reactant. H2(g)+Cl2(g)→2HCl(g)
What volume of HCl(g) measured at STP can be produced from 1.91 g of H2 and excess Cl2 according to the following equation? 16. What volume of HCl(g) measured at STP can be produced from 1.91 g of H, and excess Cl2 according to the following equation? H (8)+C12(8) → 2HCl(8) a. 21.2 L b. 42.4 L c. 86 L d. 173 L e. 64.2 L
if 250 ml of 1.50 M HCl reacts with excess zinc how many moles of zinc chloride are formed Zn(s) + 2HCl(aq) ----> ZnCl2(aq) + H2(g)
Consider the reaction: Fe + 2HCl ----> FeCl2 + H2 a) In one experiment, 0.40 moles of Fe and 0.75 moles of HCl were mixed: b) What is the limiting reactant? c) How many grams of H2 can be formed, based on the amount of the limiting reactant in this particular reaction mixture? d) How many grams of which reactant will be unreacted in this experiment?
Using the following reaction, how many moles of gallium chloride will be produced if 5.42×10^-5 moles of gallium metal react with excess hydrochloric acid? Be sure to carefully balance the equation first. Ga(s) + HCL(aq) -------> GaCl3(aq) + H2(g)
How many moles of PCl5 can be produced from 26.0 g of P4 (and excess Cl2) P4(s)+10Cl2(g)→4PCl5(g)
1. How many moles of NaCl (s) can be formed from 32 moles of Cl2 (g) reacting with an excess of Na (s)? Na (s) + Cl2 (g) -> NaCl (s) 2. How many moles of H2CO, can be formed from 2 moles of HCl reacting with an excess of Na2CO3? HCI + Na2CO3 -> NaCl + H2CO3