Answer:

Explanation:
Given that:

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

Multiplying (2) with equation (4) ; we have:

From equation (1) ; multiplying (-1) with equation (1); we have:

From equation (2); multiplying (3) with equation (2); we have:

Now; Adding up equation (5), (6) & (7) ; we get:



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
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(According to Hess Law)


1.50 mol C3H8 X (3 mol CO2 / 1 mol C3H8) X (44.0 g CO2 / 1 mol CO2) = 198 g CO2
Answer:
Zn + 2HCl → ZnCl2 + H2
Explanation:
Zn + HCl → ZnCl2 +
The complete equation is given below:
Zn+ HCl → ZnCl2 + H2
Now we can balance the equation by doing the following:
There are 2 atoms of Cl and 2 atoms of H on the left. This can be balanced by putting 2 in front of HCl as shown below:
Zn + 2HCl → ZnCl2 + H2
Answer:
(edit: nvm I figured it out, here is the answer)
Explanation: