There are 2 Ca, 4 H (2*2) and 4 O (2+2) on the left hand side. So there should be the same number of atoms 2Ca and 4(O+H) on the right hand side. The reaction is a synthesis (making something new), rather than a decomposition (destroying something).

The answer is Two molecules of Ca(OH)2 should be produced during the synthesis reaction.

4 0

4 years ago

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What is the reaction for CL^2 + 2 KBr —> 2 KCL+Br^2 of 11 grams of potassium bromide?

mylen [45]

Answer:

All the amounts of reactants and products are:

KBr

11.0g (given)

0.0924 mol

Cl₂

0.0462 mol

3.28g

KCl

0.0924 mol

6.89 g

Br₂

0.0462 mol

7.39 g

Explanation:

1. Balanced chemical equation (given)

$Cl_2+2KBr\rightarrow 2KCl+Br_2$

2. Mole ratios

$\dfrac{1molCl_2}{2molKBr}$

$\dfrac{2molKCl}{2molKBr}$

$\dfrac{1molBr_2}{2molKBr}$

3. Molar masses

Molar mass Cl₂: 70.906g/mol

Molar mass KBr: 119.002 g/mol

Molar mass KCl: 74.5513 g/mol

Molar mass KBr: 159.808 g/mol

4. Convert 11 grams of potassium bromide to moles:

#moles = mass in grams / molar mass

#mol KBr = 11g / 119.002g/mol = 0.092435mol KBr

5. Use the mole ratios to find the amounts of Cl₂, KCl, and Br₂

a) Cl₂

$\dfrac{1molCl_2}{2molKBr}\times 0.092435molKBr=0.0462molCl_2$

$0.0462175molCl_2\times 70.906g/molCl_2=3.28gCl_2$

b) KCl

$\dfrac{2molKCl}{2molKBr}\times0.092435molKBr=0.0924molKCl$

$0.092435molKCl\times 74.5513g/molKCl=6.89gKCl$

c) Br₂

$\dfrac{1molBr_2}{2molKBr}\times 0.092435molKBr=0.0462molBr_2$

$0.0462175molBr_2\times 159.808g/molBr_2=7.39gBr2$

The final calculations are rounded to 3 sginificant figures.

3 0

4 years ago