posted by Brooke
magnesium powder reacts with steam to form magnesium hydroxide and hydrogen gas. What is the pecentage yield of 10.1g Mg reacts with an excess of water abnd 21.0g Mg 9g (OH)2 is recovered?
If 24g Mg is used the percentage yield is 95% how many gramns of magnesium hydroxide should be recovered?
You REALLY should read questions you post. The typos make this almost useless; however, I will assume you meant to write, ".....and 21.0 g Mg(OH)2 is recovered?"
Mg + 2HOH ==> Mg(OH)2 + H2
1. Convert 10.1 g Mg to moles. Moles = grams/atomic mass.
2. Using the coefficients in the balanced equation, convert moles Mg to moles Mg(OH)2.
3. Now convert moles Mg(OH)2 to grams. grams = moles x molar mass.The value, in grams, which you obtain here is the THEORETICAL YIELD.
4. Now that you have the theoretical yield, you can calculate the percent yield.
%yield = (grams obtained/theoretical yield)*100 =??
For the second part, start with 24 g Mg and determine the theoretical yield. Then multiply by 0.95 to obtain the actual yield if the process is 95% efficient.
What is the percentage yield of 10.1g mg reacts with an excess of water and 21.0 g Mg(OH)2 is removed ?