THE COMPOUND IS: ethnoxybutane
Here's how to do it:
<span>Balanced equation first: </span>
<span>Mg + HCl = H2 + MgCl2 unbalanced </span>
<span>Mg + 2 HCl = H2 = MgCl balanced </span>
<span>Therefore 1 mole Mg reacts with 2 moles Hcl. </span>
<span>50g Mg = ? moles (a bit over 2; you work it out) </span>
<span>75 g HCl = ? moles (also a bit over 2; you work it out) </span>
<span>BUT, you need twice the moles HCl; therefore it is the Mg that is in excess. (you can now work out how many moles are in excess, and therefore how much mg is left over). </span>
<span>So, 2 moles HCl produce 1 mole H2(g) </span>
<span>therefore, the amount of H2 produced is half the number of moles of HCl </span>
<span>At STP, there are X litres per mole of gas (look it up - about 22 from memory) </span>
<span>Therefore, knowing the moles of H2, you can calculate the volume</span>
Answer:
396 g OF CO2 WILL BE PRODUCED BY 270 g OF GLUCOSE IN A RESPIRATION PROCESS.
Explanation:
To calculate the gram of CO2 produced by burning 270 g of gucose, we first write out the equation for the reaction and equate the two variables involved in the question;
C6H12O6 + 6O2 -------> 6CO2 + 6H2O
1 mole of C6H12O6 reacts to form 6 moles of CO2
Then, calculate the molar mass of the two variables;
Molar mass of glucose = ( 12 *6 + 1* 12 + 16* 6) g/mol = 180 g/mol
Molar mass of CO2 = (12 + 16 *2) g/mol = 44 g/mol
Next is to calculate the mass of glucose and CO2 involved in the reaction by multiplying the molar mass by the number of moles
1* 180 g of glucose yields 6 * 44 g of CO2
180 g of glucose = 264 g of CO2
If 270 g of glucose were to be used, how many grams of CO2 will be produced;
so therefore,
180 g of glucose = 264 g of CO2
270 g of glucose = x grams of CO2
x = 264 * 270 / 180
x = 71 280 / 180
x = 396 g of CO2.
In other words, 396 g of CO2 will be produced by respiration from 270 g of glucose.
