Answer:
6.50 g of Hydrogen
Explanation:
We know that in every 20.0g of sucrose, there are 1.30g of hydrogen.
We now have 100.0g of sucrose. 100.0g is 5x larger than the 20.0g sample, which is a 5 : 1 ratio. Applying this ratio to the amount of hydrogen, we would have 5*1.3g of hydrogen in the 100.0g of sucrose.
5*1.3 = 6.5, so our answer is that there are 6.50g of hydrogen in 100.0g of sucrose.
Hope this helps!
Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3
I’m positive it’s gonna be c
